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bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/ocr/mei/questions","basePath":"/a-level/maths/ocr/mei/quotient-rule/questions","formulaUrl":null,"initialIndex":16,"initialQuestionWindow":[{"id":"137c633d-888b-442a-b962-213f13a99168","revisionId":"247681ab-1766-4395-8f1e-d09549d75b04","questionNumber":null,"questionType":"LA","specification":"","difficulty":6,"difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"81485937-991c-4f60-82a8-a9496f76d288","label":null,"content":"A curve has equation \\[y = \\frac{2x + 15}{(3x - 6)^2}, \\quad x \\in \\mathbb{R}, \\quad x \\neq 2\\] Use calculus to determine the set of values of $x$ for which $y$ is increasing.","markscheme":"- Applies the quotient rule with $u = 2x + 15, v = (3x - 6)^2$, $u' = 2$ and $v' = 6(3x - 6)$ to get $\\frac{dy}{dx} = \\frac{2(3x - 6)^2 - (2x + 15)(6)(3x - 6)}{(3x - 6)^4}$ M1\n- Factorises out $(3x - 6)$ and simplifies the derivative to $\\frac{dy}{dx} = \\frac{-6x - 102}{(3x - 6)^3}$ A1\n- Sets the derivative $\\frac{dy}{dx} > 0$ and identifies the critical values $x = -17$ (where the derivative is zero) and $x = 2$ (where the derivative is undefined) M1\n- Deduces that $y$ is increasing for the range of values $-17 < x < 2$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"quotientrule","label":"Using the Quotient Rule","steps":[{"type":"text","order":0,"title":"Recall the Quotient Rule","content":"To find where the function $y$ is increasing, we need to find the derivative $\\frac{dy}{dx}$ and determine where $\\frac{dy}{dx} > 0$.\n\nSince $y$ is given as a fraction, we can write it in the form $y = \\frac{u}{v}$ where:\n$$u = 2x + 15 \\quad \\text{and} \\quad v = (3x - 6)^2$$\n\nTo differentiate $y$, we will use the **quotient rule**:\n$$\\frac{dy}{dx} = \\frac{u'v - uv'}{v^2}$$"},{"type":"text","order":1,"title":"Differentiate u and v","content":"Let's find the derivatives of $u$ and $v$ with respect to $x$:\n\n* For $u = 2x + 15$:\n$$u' = 2$$\n\n* For $v = (3x - 6)^2$, we use the **chain rule**:\n$$v' = 2(3x - 6)^1 \\cdot \\frac{d}{dx}(3x - 6) = 2(3x - 6)(3) = 6(3x - 6)$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Substitute $u$, $v$, $u'$, and $v'$ into the quotient rule formula:"},{"latex":"$$$\\frac{dy}{dx} = \\frac{\\htmlData{anchor=u-prime}{\\color{@sky}{2}}\\htmlData{anchor=v-term}{\\color{@amber}{(3x - 6)^2}} - \\htmlData{anchor=u-term}{\\color{@purple}{(2x + 15)}}\\htmlData{anchor=v-prime}{\\color{@green}{6(3x - 6)}}}{\\htmlData{anchor=v-denom}{\\color{@teal}{((3x - 6)^2)^2}}}$$"},{"latex":"$$$\\frac{dy}{dx} = \\frac{2(3x - 6)^2 - 6(2x + 15)(3x - 6)}{(3x - 6)^4}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-prime"},{"color":"amber","style":"text-color","anchor":"v-term"},{"color":"purple","style":"text-color","anchor":"u-term"},{"color":"green","style":"text-color","anchor":"v-prime"},{"color":"teal","style":"text-color","anchor":"v-denom"}],"connections":[{"to":"v-term","from":"u-prime","color":"sky","label":"u'v","style":"solid"},{"to":"v-prime","from":"u-term","color":"purple","label":"uv'","style":"solid"}]},"order":2,"title":"Apply the Quotient Rule","description":"This is our initial expression for the derivative. We will simplify it in the next step."},{"type":"text","order":3,"title":"Simplify the Derivative","content":"To simplify $\\frac{dy}{dx}$, we factorise out the common factor $(3x - 6)$ from the numerator:\n\n$$\\frac{dy}{dx} = \\frac{(3x - 6) \\left[ 2(3x - 6) - 6(2x + 15) \\right]}{(3x - 6)^4}$$\n\nCancelling $(3x - 6)$ from the numerator and denominator gives:\n\n$$\\frac{dy}{dx} = \\frac{2(3x - 6) - 6(2x + 15)}{(3x - 6)^3}$$\n\nNow, let's expand and simplify the numerator:\n\n$$2(3x - 6) - 6(2x + 15) = 6x - 12 - 12x - 90 = -6x - 102$$\n\nSubstituting this back gives our simplified derivative:\n\n$$\\frac{dy}{dx} = \\frac{-6x - 102}{(3x - 6)^3}$$"},{"type":"text","order":4,"title":"Identify Critical Values","content":"A function is increasing when its derivative is strictly positive:\n$$\\frac{dy}{dx} > 0 \\implies \\frac{-6x - 102}{(3x - 6)^3} > 0$$\n\nTo solve this inequality, we find the **critical values** where the derivative is either $0$ or undefined:\n\n1. **Numerator is zero** (where the derivative is zero):\n$$-6x - 102 = 0 \\implies -6x = 102 \\implies x = -17$$\n\n2. **Denominator is zero** (where the derivative is undefined):\n$$(3x - 6)^3 = 0 \\implies 3x - 6 = 0 \\implies x = 2$$"},{"type":"text","order":5,"title":"Perform Sign Analysis","content":"We split the real number line into three intervals using our critical values $x = -17$ and $x = 2$:\n\n* **Interval 1 ($x < -17$):** \n Let $x = -20$. \n $$\\frac{-6(-20) - 102}{(3(-20) - 6)^3} = \\frac{120 - 102}{(-66)^3} = \\frac{\\text{Positive}}{\\text{Negative}} < 0$$\n\n* **Interval 2 ($-17 < x < 2$):** \n Let $x = 0$. \n $$\\frac{-6(0) - 102}{(3(0) - 6)^3} = \\frac{-102}{(-6)^3} = \\frac{\\text{Negative}}{\\text{Negative}} > 0$$\n\n* **Interval 3 ($x > 2$):** \n Let $x = 3$. \n $$\\frac{-6(3) - 102}{(3(3) - 6)^3} = \\frac{-120}{3^3} = \\frac{\\text{Negative}}{\\text{Positive}} < 0$$\n\nThus, the derivative is positive only on the interval $-17 < x < 2$. Since $x \\neq 2$ is already specified, the set of values for which $y$ is increasing is:\n$$-17 < x < 2$$"}]},{"id":"productrule","label":"Using the Product Rule","steps":[{"type":"text","order":0,"title":"Rewrite as a product","content":"To differentiate the function using the product rule, we first rewrite the fraction as a product with a negative exponent:\n\n$$y = (2x + 15)(3x - 6)^{-2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u-def}{\\color{@sky}{u = 2x + 15}} \\quad \\text{and} \\quad \\htmlData{anchor=v-def}{\\color{@purple}{v = (3x - 6)^{-2}}}$"},{"latex":"$$\\text{Differentiating both parts: } \\htmlData{anchor=du-def}{\\color{@sky}{u' = 2}} \\quad \\text{and} \\quad \\htmlData{anchor=dv-def}{\\color{@purple}{v' = -6(3x - 6)^{-3}}}$"},{"latex":"$$$\\frac{\\text{d}y}{\\text{d}x} = \\htmlData{anchor=u-term}{\\color{@sky}{u}}\\htmlData{anchor=dv-term}{\\color{@purple}{v'}} + \\htmlData{anchor=v-term}{\\color{@purple}{v}}\\htmlData{anchor=du-term}{\\color{@sky}{u'}}$$"},{"text":"Substituting these expressions gives the derivative before simplifying:"},{"latex":"$$$\\frac{\\text{d}y}{\\text{d}x} = \\color{@sky}{(2x + 15)}\\color{@purple}{\\left[-6(3x - 6)^{-3}\\right]} + \\color{@purple}{(3x - 6)^{-2}}\\color{@sky}{(2)}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"purple","style":"text-color","anchor":"v-def"},{"color":"sky","style":"text-color","anchor":"du-def"},{"color":"purple","style":"text-color","anchor":"dv-def"}],"connections":[{"to":"u-term","from":"u-def","color":"sky","label":"u","style":"solid"},{"to":"v-term","from":"v-def","color":"purple","label":"v","style":"solid"},{"to":"du-term","from":"du-def","color":"sky","label":"u'","style":"solid"},{"to":"dv-term","from":"dv-def","color":"purple","label":"v'","style":"solid"}]},"order":1,"title":"Apply the product rule","description":"We identify $u$ and $v$, use the chain rule to find $v'$, and substitute them into the product rule formula."},{"type":"text","order":2,"title":"Factorise and simplify","content":"To simplify, we factor out the lowest power of $(3x - 6)$, which is $(3x - 6)^{-3}$:\n\n$$\\frac{\\text{d}y}{\\text{d}x} = (3x - 6)^{-3} \\left[ -6(2x + 15) + 2(3x - 6) \\right]$$\n\nNow we simplify the terms inside the square brackets:\n\n$$-6(2x + 15) + 2(3x - 6) = -12x - 90 + 6x - 12 = -6x - 102$$\n\nSubstituting this back yields:\n\n$$\\frac{\\text{d}y}{\\text{d}x} = (3x - 6)^{-3} [-6x - 102]$$\n\nRewriting this back as a single fraction gives:\n\n$$\\frac{\\text{d}y}{\\text{d}x} = \\frac{-6x - 102}{(3x - 6)^3}$$"},{"type":"text","order":3,"title":"Set up the inequality","content":"A function is increasing when its derivative is strictly positive:\n\n$$\\frac{\\text{d}y}{\\text{d}x} > 0 \\implies \\frac{-6x - 102}{(3x - 6)^3} > 0$$"},{"type":"text","order":4,"title":"Solve for x","content":"To find the set of values where this inequality holds, we identify the critical boundaries:\n\n* The numerator is zero when:\n$$-6x - 102 = 0 \\implies x = -17$$\n\n* The denominator is zero when:\n$$3x - 6 = 0 \\implies x = 2$$\n\nNow, we test the sign of $\\frac{\\text{d}y}{\\text{d}x}$ in the intervals created by these boundary points:\n\n* For $x < -17$: the numerator is positive and the denominator is negative, so $\\frac{\\text{d}y}{\\text{d}x} < 0$.\n* For $-17 < x < 2$: the numerator is negative and the denominator is negative, so $\\frac{\\text{d}y}{\\text{d}x} > 0$.\n* For $x > 2$: the numerator is negative and the denominator is positive, so $\\frac{\\text{d}y}{\\text{d}x} < 0$.\n\nTherefore, the set of values for which $y$ is increasing is:\n\n$$-17 < x < 2$$"}]}],"generatedAt":"2026-06-09T03:26:17.228Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"81485937-991c-4f60-82a8-a9496f76d288:quotientrule","label":"Using the Quotient Rule","steps":[{"type":"text","order":0,"title":"Recall the Quotient Rule","content":"To find where the function $y$ is increasing, we need to find the derivative $\\frac{dy}{dx}$ and determine where $\\frac{dy}{dx} > 0$.\n\nSince $y$ is given as a fraction, we can write it in the form $y = \\frac{u}{v}$ where:\n$$u = 2x + 15 \\quad \\text{and} \\quad v = (3x - 6)^2$$\n\nTo differentiate $y$, we will use the **quotient rule**:\n$$\\frac{dy}{dx} = \\frac{u'v - uv'}{v^2}$$"},{"type":"text","order":1,"title":"Differentiate u and v","content":"Let's find the derivatives of $u$ and $v$ with respect to $x$:\n\n* For $u = 2x + 15$:\n$$u' = 2$$\n\n* For $v = (3x - 6)^2$, we use the **chain rule**:\n$$v' = 2(3x - 6)^1 \\cdot \\frac{d}{dx}(3x - 6) = 2(3x - 6)(3) = 6(3x - 6)$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Substitute $u$, $v$, $u'$, and $v'$ into the quotient rule formula:"},{"latex":"$$$\\frac{dy}{dx} = \\frac{\\htmlData{anchor=u-prime}{\\color{@sky}{2}}\\htmlData{anchor=v-term}{\\color{@amber}{(3x - 6)^2}} - \\htmlData{anchor=u-term}{\\color{@purple}{(2x + 15)}}\\htmlData{anchor=v-prime}{\\color{@green}{6(3x - 6)}}}{\\htmlData{anchor=v-denom}{\\color{@teal}{((3x - 6)^2)^2}}}$$"},{"latex":"$$$\\frac{dy}{dx} = \\frac{2(3x - 6)^2 - 6(2x + 15)(3x - 6)}{(3x - 6)^4}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-prime"},{"color":"amber","style":"text-color","anchor":"v-term"},{"color":"purple","style":"text-color","anchor":"u-term"},{"color":"green","style":"text-color","anchor":"v-prime"},{"color":"teal","style":"text-color","anchor":"v-denom"}],"connections":[{"to":"v-term","from":"u-prime","color":"sky","label":"u'v","style":"solid"},{"to":"v-prime","from":"u-term","color":"purple","label":"uv'","style":"solid"}]},"order":2,"title":"Apply the Quotient Rule","description":"This is our initial expression for the derivative. We will simplify it in the next step."},{"type":"text","order":3,"title":"Simplify the Derivative","content":"To simplify $\\frac{dy}{dx}$, we factorise out the common factor $(3x - 6)$ from the numerator:\n\n$$\\frac{dy}{dx} = \\frac{(3x - 6) \\left[ 2(3x - 6) - 6(2x + 15) \\right]}{(3x - 6)^4}$$\n\nCancelling $(3x - 6)$ from the numerator and denominator gives:\n\n$$\\frac{dy}{dx} = \\frac{2(3x - 6) - 6(2x + 15)}{(3x - 6)^3}$$\n\nNow, let's expand and simplify the numerator:\n\n$$2(3x - 6) - 6(2x + 15) = 6x - 12 - 12x - 90 = -6x - 102$$\n\nSubstituting this back gives our simplified derivative:\n\n$$\\frac{dy}{dx} = \\frac{-6x - 102}{(3x - 6)^3}$$"},{"type":"text","order":4,"title":"Identify Critical Values","content":"A function is increasing when its derivative is strictly positive:\n$$\\frac{dy}{dx} > 0 \\implies \\frac{-6x - 102}{(3x - 6)^3} > 0$$\n\nTo solve this inequality, we find the **critical values** where the derivative is either $0$ or undefined:\n\n1. **Numerator is zero** (where the derivative is zero):\n$$-6x - 102 = 0 \\implies -6x = 102 \\implies x = -17$$\n\n2. **Denominator is zero** (where the derivative is undefined):\n$$(3x - 6)^3 = 0 \\implies 3x - 6 = 0 \\implies x = 2$$"},{"type":"text","order":5,"title":"Perform Sign Analysis","content":"We split the real number line into three intervals using our critical values $x = -17$ and $x = 2$:\n\n* **Interval 1 ($x < -17$):** \n Let $x = -20$. \n $$\\frac{-6(-20) - 102}{(3(-20) - 6)^3} = \\frac{120 - 102}{(-66)^3} = \\frac{\\text{Positive}}{\\text{Negative}} < 0$$\n\n* **Interval 2 ($-17 < x < 2$):** \n Let $x = 0$. \n $$\\frac{-6(0) - 102}{(3(0) - 6)^3} = \\frac{-102}{(-6)^3} = \\frac{\\text{Negative}}{\\text{Negative}} > 0$$\n\n* **Interval 3 ($x > 2$):** \n Let $x = 3$. \n $$\\frac{-6(3) - 102}{(3(3) - 6)^3} = \\frac{-120}{3^3} = \\frac{\\text{Negative}}{\\text{Positive}} < 0$$\n\nThus, the derivative is positive only on the interval $-17 < x < 2$. Since $x \\neq 2$ is already specified, the set of values for which $y$ is increasing is:\n$$-17 < x < 2$$"}]},{"id":"81485937-991c-4f60-82a8-a9496f76d288:productrule","label":"Using the Product Rule","steps":[{"type":"text","order":0,"title":"Rewrite as a product","content":"To differentiate the function using the product rule, we first rewrite the fraction as a product with a negative exponent:\n\n$$y = (2x + 15)(3x - 6)^{-2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u-def}{\\color{@sky}{u = 2x + 15}} \\quad \\text{and} \\quad \\htmlData{anchor=v-def}{\\color{@purple}{v = (3x - 6)^{-2}}}$"},{"latex":"$$\\text{Differentiating both parts: } \\htmlData{anchor=du-def}{\\color{@sky}{u' = 2}} \\quad \\text{and} \\quad \\htmlData{anchor=dv-def}{\\color{@purple}{v' = -6(3x - 6)^{-3}}}$"},{"latex":"$$$\\frac{\\text{d}y}{\\text{d}x} = \\htmlData{anchor=u-term}{\\color{@sky}{u}}\\htmlData{anchor=dv-term}{\\color{@purple}{v'}} + \\htmlData{anchor=v-term}{\\color{@purple}{v}}\\htmlData{anchor=du-term}{\\color{@sky}{u'}}$$"},{"text":"Substituting these expressions gives the derivative before simplifying:"},{"latex":"$$$\\frac{\\text{d}y}{\\text{d}x} = \\color{@sky}{(2x + 15)}\\color{@purple}{\\left[-6(3x - 6)^{-3}\\right]} + \\color{@purple}{(3x - 6)^{-2}}\\color{@sky}{(2)}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"purple","style":"text-color","anchor":"v-def"},{"color":"sky","style":"text-color","anchor":"du-def"},{"color":"purple","style":"text-color","anchor":"dv-def"}],"connections":[{"to":"u-term","from":"u-def","color":"sky","label":"u","style":"solid"},{"to":"v-term","from":"v-def","color":"purple","label":"v","style":"solid"},{"to":"du-term","from":"du-def","color":"sky","label":"u'","style":"solid"},{"to":"dv-term","from":"dv-def","color":"purple","label":"v'","style":"solid"}]},"order":1,"title":"Apply the product rule","description":"We identify $u$ and $v$, use the chain rule to find $v'$, and substitute them into the product rule formula."},{"type":"text","order":2,"title":"Factorise and simplify","content":"To simplify, we factor out the lowest power of $(3x - 6)$, which is $(3x - 6)^{-3}$:\n\n$$\\frac{\\text{d}y}{\\text{d}x} = (3x - 6)^{-3} \\left[ -6(2x + 15) + 2(3x - 6) \\right]$$\n\nNow we simplify the terms inside the square brackets:\n\n$$-6(2x + 15) + 2(3x - 6) = -12x - 90 + 6x - 12 = -6x - 102$$\n\nSubstituting this back yields:\n\n$$\\frac{\\text{d}y}{\\text{d}x} = (3x - 6)^{-3} [-6x - 102]$$\n\nRewriting this back as a single fraction gives:\n\n$$\\frac{\\text{d}y}{\\text{d}x} = \\frac{-6x - 102}{(3x - 6)^3}$$"},{"type":"text","order":3,"title":"Set up the inequality","content":"A function is increasing when its derivative is strictly positive:\n\n$$\\frac{\\text{d}y}{\\text{d}x} > 0 \\implies \\frac{-6x - 102}{(3x - 6)^3} > 0$$"},{"type":"text","order":4,"title":"Solve for x","content":"To find the set of values where this inequality holds, we identify the critical boundaries:\n\n* The numerator is zero when:\n$$-6x - 102 = 0 \\implies x = -17$$\n\n* The denominator is zero when:\n$$3x - 6 = 0 \\implies x = 2$$\n\nNow, we test the sign of $\\frac{\\text{d}y}{\\text{d}x}$ in the intervals created by these boundary points:\n\n* For $x < -17$: the numerator is positive and the denominator is negative, so $\\frac{\\text{d}y}{\\text{d}x} < 0$.\n* For $-17 < x < 2$: the numerator is negative and the denominator is negative, so $\\frac{\\text{d}y}{\\text{d}x} > 0$.\n* For $x > 2$: the numerator is negative and the denominator is positive, so $\\frac{\\text{d}y}{\\text{d}x} < 0$.\n\nTherefore, the set of values for which $y$ is increasing is:\n\n$$-17 < x < 2$$"}]}],"generatedAt":"2026-06-09T03:26:37.461Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"1dd6e7ed-234f-4728-8cdb-9ab46caf8736","revisionId":"c841e1cc-5490-49c4-86fd-bbea13a06cd3","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"4f67d304-57e3-4673-b38c-baa454497e28","label":null,"content":"Given that \\[y = \\frac{x^4}{\\cos x}\\] find \$\\frac{\\text{d}y}{\\text{d}x}\$.","markscheme":"- Identifies $u = x^4 \\Rightarrow \\frac{du}{dx} = 4x^3$ and $v = \\cos x \\Rightarrow \\frac{dv}{dx} = -\\sin x$ M1\n- Substitutes these derivatives into the quotient rule formula to obtain $\\frac{dy}{dx} = \\frac{(\\cos x)(4x^3) - (x^4)(-\\sin x)}{(\\cos x)^2}$ M1\n- Simplifies the numerator to reach the final expression $\\frac{4x^3 \\cos x + x^4 \\sin x}{\\cos^2 x}$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"quotient_rule","label":"Quotient Rule","steps":[{"type":"text","order":0,"title":"Identify the quotient rule","content":"We need to find the derivative of the function:\n\n$$y = \\frac{x^4}{\\cos x}$$\n\nSince this function is in the form of a fraction where both the numerator and denominator are functions of $x$, we use the **Quotient Rule**. From the data booklet, the quotient rule is:\n\n$$\\text{If } y = \\frac{u}{v}, \\text{ then } \\frac{\\text{d}y}{\\text{d}x} = \\frac{v\\frac{\\text{d}u}{\\text{d}x} - u\\frac{\\text{d}v}{\\text{d}x}}{v^2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{\\htmlData{anchor=u-expr}{\\color{@sky}{x^4}}}{\\htmlData{anchor=v-expr}{\\color{@amber}{\\cos x}}}$"},{"latex":"$$\\htmlData{anchor=u-def}{\\color{@sky}{u = x^4}} \\implies \\htmlData{anchor=du-def}{\\color{@purple}{\\frac{\\text{d}u}{\\text{d}x} = 4x^3}}$"},{"latex":"$$\\htmlData{anchor=v-def}{\\color{@amber}{v = \\cos x}} \\implies \\htmlData{anchor=dv-def}{\\color{@teal}{\\frac{\\text{d}v}{\\text{d}x} = -\\sin x}}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"u-def"},{"color":"amber","style":"underline","anchor":"v-def"}],"connections":[{"to":"u-def","from":"u-expr","color":"sky","label":null,"style":"solid"},{"to":"v-def","from":"v-expr","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Define $u$ and $v$","description":"Identify the numerator $u$ and the denominator $v$, and differentiate each with respect to $x$ to find $\\frac{\\text{d}u}{\\text{d}x}$ and $\\frac{\\text{d}v}{\\text{d}x}$. This step earns you the first mark (**M1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\text{d}y}{\\text{d}x} = \\frac{\\htmlData{anchor=v-sub-eq}{\\color{@amber}{v}} \\cdot \\htmlData{anchor=du-sub-eq}{\\color{@purple}{\\frac{\\text{d}u}{\\text{d}x}}} - \\htmlData{anchor=u-sub-eq}{\\color{@sky}{u}} \\cdot \\htmlData{anchor=dv-sub-eq}{\\color{@teal}{\\frac{\\text{d}v}{\\text{d}x}}}}{\\htmlData{anchor=v2-sub-eq}{\\color{@amber}{v^2}}}$"},{"latex":"$$\\phantom{\\frac{\\text{d}y}{\\text{d}x}} = \\frac{(\\htmlData{anchor=v-val}{\\color{@amber}{\\cos x}})(\\htmlData{anchor=du-val}{\\color{@purple}{4x^3}}) - (\\htmlData{anchor=u-val}{\\color{@sky}{x^4}})(\\htmlData{anchor=dv-val}{\\color{@teal}{-\\sin x}})}{(\\htmlData{anchor=v2-val}{\\color{@amber}{\\cos x}})^2}$"}],"highlights":[{"color":"amber","style":"underline","anchor":"v-val"},{"color":"purple","style":"underline","anchor":"du-val"}],"connections":[{"to":"v-val","from":"v-sub-eq","color":"amber","label":null,"style":"solid"},{"to":"du-val","from":"du-sub-eq","color":"purple","label":null,"style":"solid"},{"to":"u-val","from":"u-sub-eq","color":"sky","label":null,"style":"solid"},{"to":"dv-val","from":"dv-sub-eq","color":"teal","label":null,"style":"solid"}]},"order":2,"title":"Apply quotient rule","description":"Substitute $u$, $v$, $\\frac{\\text{d}u}{\\text{d}x}$, and $\\frac{\\text{d}v}{\\text{d}x}$ into the quotient rule formula. This substitution earns you the second mark (**M1**)."},{"type":"text","order":3,"title":"Simplify final expression","content":"Now, we simplify the terms in the numerator:\n\n* The first term is $(\\cos x)(4x^3) = 4x^3 \\cos x$\n* The second term is $- (x^4)(-\\sin x) = + x^4 \\sin x$\n* The denominator is $(\\cos x)^2 = \\cos^2 x$\n\nPutting it all together, we get our final simplified answer:\n\n$$\\frac{\\text{d}y}{\\text{d}x} = \\frac{4x^3 \\cos x + x^4 \\sin x}{\\cos^2 x}$$\n\nThis final simplification earns you the accuracy mark (**A1**)."}]},{"id":"product_rule","label":"Product Rule","steps":[{"type":"text","order":0,"title":"Rewrite as a product","content":"To differentiate the given function using the product rule, we first rewrite the quotient as a product of two terms: $$y = x^4 (\\cos x)^{-1}$$ Now, we can define our two functions: $$u = x^4$$ and $$v = (\\cos x)^{-1}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = x^4 \\implies \\htmlData{anchor=du-val}{\\frac{du}{dx} = 4x^3}$"},{"latex":"$$v = (\\cos x)^{-1} \\implies \\htmlData{anchor=dv-val}{\\frac{dv}{dx} = -1(\\cos x)^{-2} \\cdot (-\\sin x) = \\frac{\\sin x}{\\cos^2 x}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"du-val"},{"color":"pink","style":"box","anchor":"dv-val"}],"connections":null},"order":1,"title":"Find the individual derivatives","description":"We differentiate $u$ normally using the power rule, and we differentiate $v$ using the chain rule."},{"type":"text","order":2,"title":"Apply the product rule","content":"The product rule states that: $$\\frac{dy}{dx} = u\\frac{dv}{dx} + v\\frac{du}{dx}$$ Substituting our functions and their derivatives into this formula, we get: $$\\frac{dy}{dx} = (x^4)\\left(\\frac{\\sin x}{\\cos^2 x}\\right) + (\\cos x)^{-1}(4x^3)$$ This simplifies to: $$\\frac{dy}{dx} = \\frac{x^4 \\sin x}{\\cos^2 x} + \\frac{4x^3}{\\cos x}$$"},{"type":"text","order":3,"title":"Simplify with common denominator","content":"To obtain the final simplified expression, we express both terms over a common denominator of $\\cos^2 x$. Multiplying the numerator and denominator of the second term by $\\cos x$ gives: $$\\frac{dy}{dx} = \\frac{x^4 \\sin x}{\\cos^2 x} + \\frac{4x^3 \\cos x}{\\cos^2 x}$$ Combining these over the single denominator yields the final answer: $$\\frac{dy}{dx} = \\frac{4x^3 \\cos x + x^4 \\sin x}{\\cos^2 x}$$"}]},{"id":"logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Introduce Logarithmic Differentiation","content":"To differentiate $y = \\frac{x^4}{\\cos x}$ using logarithmic differentiation, we begin by taking the natural logarithm ($\\ln$) of both sides of the equation:\n\n$$\\ln(y) = \\ln\\left(\\frac{x^4}{\\cos x}\\right)$$\n\nThis method is particularly useful here as it allows us to convert the quotient into simpler, separate terms before differentiating."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\ln(y) = \\htmlData{anchor=log-frac}{\\ln\\left(\\frac{x^4}{\\cos x}\\right)}$"},{"latex":"$$\\ln(y) = \\htmlData{anchor=term1}{\\ln(x^4)} - \\htmlData{anchor=term2}{\\ln(\\cos x)}$"},{"latex":"$$\\ln(y) = \\htmlData{anchor=final-exp}{4\\ln(x)} - \\ln(\\cos x)$"}],"highlights":[{"color":"sky","style":"box","anchor":"log-frac"},{"color":"sky","style":"text-color","anchor":"term1"},{"color":"pink","style":"text-color","anchor":"term2"},{"color":"amber","style":"text-color","anchor":"final-exp"}],"connections":[{"to":"term1","from":"log-frac","color":"sky","label":"log laws","style":"solid"},{"to":"final-exp","from":"term1","color":"amber","label":"power rule","style":"solid"}]},"order":1,"title":"Expand using log laws","description":"We use the logarithm laws $\\ln\\left(\\frac{a}{b}\\right) = \\ln(a) - \\ln(b)$ and $\\ln(a^k) = k\\ln(a)$ to expand the right-hand side."},{"type":"text","order":2,"title":"Differentiate implicitly","content":"Now we differentiate both sides of the equation with respect to $x$.\n\nOn the left-hand side, we apply the chain rule (implicit differentiation):\n$$\\frac{d}{dx}[\\ln(y)] = \\frac{1}{y}\\frac{dy}{dx}$$\n\nOn the right-hand side, we differentiate term-by-term:\n- The derivative of $4\\ln(x)$ is $\\frac{4}{x}$.\n- For $\\ln(\\cos x)$, we use the chain rule:\n $$\\frac{d}{dx}[\\ln(\\cos x)] = \\frac{1}{\\cos x} \\cdot (-\\sin x) = -\\tan x$$\n\nPutting this together, we get:\n$$\\frac{1}{y}\\frac{dy}{dx} = \\frac{4}{x} - (-\\tan x) = \\frac{4}{x} + \\frac{\\sin x}{\\cos x}$$"},{"type":"text","order":3,"title":"Solve for dy/dx","content":"Multiply both sides of the equation by $y$ to isolate $\\frac{dy}{dx}$:\n\n$$\\frac{dy}{dx} = y \\left(\\frac{4}{x} + \\frac{\\sin x}{\\cos x}\\right)$$\n\nNow, substitute the original expression for $y = \\frac{x^4}{\\cos x}$ back into this equation:\n\n$$\\frac{dy}{dx} = \\left(\\frac{x^4}{\\cos x}\\right)\\left(\\frac{4}{x} + \\frac{\\sin x}{\\cos x}\\right)$$\n\nExpanding the brackets gives:\n\n$$\\frac{dy}{dx} = \\frac{4x^3}{\\cos x} + \\frac{x^4\\sin x}{\\cos^2 x}$$"},{"type":"text","order":4,"title":"Simplify the final fraction","content":"To match standard forms, we write the terms over a common denominator of $\\cos^2 x$:\n\n$$\\frac{dy}{dx} = \\frac{4x^3\\cos x}{\\cos^2 x} + \\frac{x^4\\sin x}{\\cos^2 x}$$\n\nCombining these into a single fraction gives the final derivative:\n\n$$\\frac{dy}{dx} = \\frac{4x^3\\cos x + x^4\\sin x}{\\cos^2 x}$$"}]}],"generatedAt":"2026-06-09T03:42:27.203Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"4f67d304-57e3-4673-b38c-baa454497e28:quotient_rule","label":"Quotient Rule","steps":[{"type":"text","order":0,"title":"Identify the quotient rule","content":"We need to find the derivative of the function:\n\n$$y = \\frac{x^4}{\\cos x}$$\n\nSince this function is in the form of a fraction where both the numerator and denominator are functions of $x$, we use the **Quotient Rule**. From the data booklet, the quotient rule is:\n\n$$\\text{If } y = \\frac{u}{v}, \\text{ then } \\frac{\\text{d}y}{\\text{d}x} = \\frac{v\\frac{\\text{d}u}{\\text{d}x} - u\\frac{\\text{d}v}{\\text{d}x}}{v^2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{\\htmlData{anchor=u-expr}{\\color{@sky}{x^4}}}{\\htmlData{anchor=v-expr}{\\color{@amber}{\\cos x}}}$"},{"latex":"$$\\htmlData{anchor=u-def}{\\color{@sky}{u = x^4}} \\implies \\htmlData{anchor=du-def}{\\color{@purple}{\\frac{\\text{d}u}{\\text{d}x} = 4x^3}}$"},{"latex":"$$\\htmlData{anchor=v-def}{\\color{@amber}{v = \\cos x}} \\implies \\htmlData{anchor=dv-def}{\\color{@teal}{\\frac{\\text{d}v}{\\text{d}x} = -\\sin x}}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"u-def"},{"color":"amber","style":"underline","anchor":"v-def"}],"connections":[{"to":"u-def","from":"u-expr","color":"sky","label":null,"style":"solid"},{"to":"v-def","from":"v-expr","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Define $u$ and $v$","description":"Identify the numerator $u$ and the denominator $v$, and differentiate each with respect to $x$ to find $\\frac{\\text{d}u}{\\text{d}x}$ and $\\frac{\\text{d}v}{\\text{d}x}$. This step earns you the first mark (**M1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\text{d}y}{\\text{d}x} = \\frac{\\htmlData{anchor=v-sub-eq}{\\color{@amber}{v}} \\cdot \\htmlData{anchor=du-sub-eq}{\\color{@purple}{\\frac{\\text{d}u}{\\text{d}x}}} - \\htmlData{anchor=u-sub-eq}{\\color{@sky}{u}} \\cdot \\htmlData{anchor=dv-sub-eq}{\\color{@teal}{\\frac{\\text{d}v}{\\text{d}x}}}}{\\htmlData{anchor=v2-sub-eq}{\\color{@amber}{v^2}}}$"},{"latex":"$$\\phantom{\\frac{\\text{d}y}{\\text{d}x}} = \\frac{(\\htmlData{anchor=v-val}{\\color{@amber}{\\cos x}})(\\htmlData{anchor=du-val}{\\color{@purple}{4x^3}}) - (\\htmlData{anchor=u-val}{\\color{@sky}{x^4}})(\\htmlData{anchor=dv-val}{\\color{@teal}{-\\sin x}})}{(\\htmlData{anchor=v2-val}{\\color{@amber}{\\cos x}})^2}$"}],"highlights":[{"color":"amber","style":"underline","anchor":"v-val"},{"color":"purple","style":"underline","anchor":"du-val"}],"connections":[{"to":"v-val","from":"v-sub-eq","color":"amber","label":null,"style":"solid"},{"to":"du-val","from":"du-sub-eq","color":"purple","label":null,"style":"solid"},{"to":"u-val","from":"u-sub-eq","color":"sky","label":null,"style":"solid"},{"to":"dv-val","from":"dv-sub-eq","color":"teal","label":null,"style":"solid"}]},"order":2,"title":"Apply quotient rule","description":"Substitute $u$, $v$, $\\frac{\\text{d}u}{\\text{d}x}$, and $\\frac{\\text{d}v}{\\text{d}x}$ into the quotient rule formula. This substitution earns you the second mark (**M1**)."},{"type":"text","order":3,"title":"Simplify final expression","content":"Now, we simplify the terms in the numerator:\n\n* The first term is $(\\cos x)(4x^3) = 4x^3 \\cos x$\n* The second term is $- (x^4)(-\\sin x) = + x^4 \\sin x$\n* The denominator is $(\\cos x)^2 = \\cos^2 x$\n\nPutting it all together, we get our final simplified answer:\n\n$$\\frac{\\text{d}y}{\\text{d}x} = \\frac{4x^3 \\cos x + x^4 \\sin x}{\\cos^2 x}$$\n\nThis final simplification earns you the accuracy mark (**A1**)."}]},{"id":"4f67d304-57e3-4673-b38c-baa454497e28:product_rule","label":"Product Rule","steps":[{"type":"text","order":0,"title":"Rewrite as a product","content":"To differentiate the given function using the product rule, we first rewrite the quotient as a product of two terms: $$y = x^4 (\\cos x)^{-1}$$ Now, we can define our two functions: $$u = x^4$$ and $$v = (\\cos x)^{-1}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = x^4 \\implies \\htmlData{anchor=du-val}{\\frac{du}{dx} = 4x^3}$"},{"latex":"$$v = (\\cos x)^{-1} \\implies \\htmlData{anchor=dv-val}{\\frac{dv}{dx} = -1(\\cos x)^{-2} \\cdot (-\\sin x) = \\frac{\\sin x}{\\cos^2 x}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"du-val"},{"color":"pink","style":"box","anchor":"dv-val"}],"connections":null},"order":1,"title":"Find the individual derivatives","description":"We differentiate $u$ normally using the power rule, and we differentiate $v$ using the chain rule."},{"type":"text","order":2,"title":"Apply the product rule","content":"The product rule states that: $$\\frac{dy}{dx} = u\\frac{dv}{dx} + v\\frac{du}{dx}$$ Substituting our functions and their derivatives into this formula, we get: $$\\frac{dy}{dx} = (x^4)\\left(\\frac{\\sin x}{\\cos^2 x}\\right) + (\\cos x)^{-1}(4x^3)$$ This simplifies to: $$\\frac{dy}{dx} = \\frac{x^4 \\sin x}{\\cos^2 x} + \\frac{4x^3}{\\cos x}$$"},{"type":"text","order":3,"title":"Simplify with common denominator","content":"To obtain the final simplified expression, we express both terms over a common denominator of $\\cos^2 x$. Multiplying the numerator and denominator of the second term by $\\cos x$ gives: $$\\frac{dy}{dx} = \\frac{x^4 \\sin x}{\\cos^2 x} + \\frac{4x^3 \\cos x}{\\cos^2 x}$$ Combining these over the single denominator yields the final answer: $$\\frac{dy}{dx} = \\frac{4x^3 \\cos x + x^4 \\sin x}{\\cos^2 x}$$"}]},{"id":"4f67d304-57e3-4673-b38c-baa454497e28:logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Introduce Logarithmic Differentiation","content":"To differentiate $y = \\frac{x^4}{\\cos x}$ using logarithmic differentiation, we begin by taking the natural logarithm of the absolute value of both sides. For $x \\ne 0$ and $\\cos x \\ne 0$, we have $y \\ne 0$, so:\n\n$$\\ln|y| = \\ln\\left|\\frac{x^4}{\\cos x}\\right|$$\n\nThe absolute values ensure that the logarithms are defined when $x$ or $y$ is negative. The case $x=0$, which is excluded by this method, will be checked separately."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\ln|y| = \\htmlData{anchor=log-frac}{\\ln\\left|\\frac{x^4}{\\cos x}\\right|}"},{"latex":"\\ln|y| = \\htmlData{anchor=term1}{\\ln(x^4)} - \\htmlData{anchor=term2}{\\ln|\\cos x|}"},{"latex":"\\ln|y| = \\htmlData{anchor=final-exp}{4\\ln|x|} - \\ln|\\cos x|"}],"highlights":[{"color":"sky","style":"box","anchor":"log-frac"},{"color":"sky","style":"text-color","anchor":"term1"},{"color":"pink","style":"text-color","anchor":"term2"},{"color":"amber","style":"text-color","anchor":"final-exp"}],"connections":[{"to":"term1","from":"log-frac","color":"sky","label":"log laws","style":"solid"},{"to":"final-exp","from":"term1","color":"amber","label":"power rule","style":"solid"}]},"order":1,"title":"Expand using log laws","description":"We use the logarithm laws $\\ln\\left|\\frac{a}{b}\\right| = \\ln|a| - \\ln|b|$ and, for $x \\ne 0$, $\\ln(x^4) = 4\\ln|x|$ to expand the right-hand side."},{"type":"text","order":2,"title":"Differentiate implicitly","content":"Now we differentiate both sides of the equation with respect to $x$, for $x \\ne 0$ and $\\cos x \\ne 0$.\n\nOn the left-hand side, we apply the chain rule (implicit differentiation):\n$$\\frac{d}{dx}[\\ln|y|] = \\frac{1}{y}\\frac{dy}{dx}$$\n\nOn the right-hand side, we differentiate term-by-term:\n- The derivative of $4\\ln|x|$ is $\\frac{4}{x}$.\n- For $\\ln|\\cos x|$, we use the chain rule:\n $$\\frac{d}{dx}[\\ln|\\cos x|] = \\frac{1}{\\cos x} \\cdot (-\\sin x) = -\\tan x$$\n\nPutting this together, we get:\n$$\\frac{1}{y}\\frac{dy}{dx} = \\frac{4}{x} - (-\\tan x) = \\frac{4}{x} + \\frac{\\sin x}{\\cos x}$$"},{"type":"text","order":3,"title":"Solve for dy/dx","content":"For $x \\ne 0$ and $\\cos x \\ne 0$, multiply both sides of the equation by $y$ to isolate $\\frac{dy}{dx}$:\n\n$$\\frac{dy}{dx} = y \\left(\\frac{4}{x} + \\frac{\\sin x}{\\cos x}\\right)$$\n\nNow, substitute the original expression for $y = \\frac{x^4}{\\cos x}$ back into this equation:\n\n$$\\frac{dy}{dx} = \\left(\\frac{x^4}{\\cos x}\\right)\\left(\\frac{4}{x} + \\frac{\\sin x}{\\cos x}\\right)$$\n\nExpanding the brackets gives:\n\n$$\\frac{dy}{dx} = \\frac{4x^3}{\\cos x} + \\frac{x^4\\sin x}{\\cos^2 x}$$"},{"type":"text","order":4,"title":"Simplify the final fraction","content":"For $x \\ne 0$ and $\\cos x \\ne 0$, we write the terms over a common denominator of $\\cos^2 x$:\n\n$$\\frac{dy}{dx} = \\frac{4x^3\\cos x}{\\cos^2 x} + \\frac{x^4\\sin x}{\\cos^2 x}$$\n\nCombining these into a single fraction gives:\n\n$$\\frac{dy}{dx} = \\frac{4x^3\\cos x + x^4\\sin x}{\\cos^2 x}$$\n\nIt remains to check $x=0$, since logarithmic differentiation excluded this value. From $y=x^4\\sec x$,\n$$y'(0)=\\lim_{h\\to 0}\\frac{h^4\\sec h}{h}=\\lim_{h\\to 0}h^3\\sec h=0.$$\nThe final expression also gives $0$ when $x=0$, so the result is valid for every $x$ such that $\\cos x\\ne 0$."}]}],"generatedAt":"2026-06-09T03:42:39.545Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"362274ef-ce53-4c09-a816-30ac6ec89d5d","revisionId":"ad58509a-49ea-4569-85af-3f04c060deb6","questionNumber":null,"questionType":"LA","specification":"A chemical engineer monitors the concentration, $C$ mmol/L, of a reagent in a reaction vessel, which is modeled by the function\n\n$$C(t) = \\frac{4t + 8}{\\sqrt{2t + 1}}, \\quad t > -0.5$$\n\nwhere $t$ is the time in minutes from the start of the reaction.","difficulty":5,"difficultyBand":"medium","tier":null,"totalMarks":7,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"7e86b414-ab24-4f77-bc77-70398962ff40","label":"a","content":"Find $C'(t)$, giving your answer in simplest form.","markscheme":"- Identifies $u = 4t + 8$ and $v = (2t + 1)^{1/2}$ and applies the quotient rule $\\frac{v u' - u v'}{v^2}$ M1\n- Obtains $u' = 4$ and uses the chain rule to find $v' = \\frac{1}{2}(2t + 1)^{-1/2} \\times 2 = (2t + 1)^{-1/2}$ A1\n- Substitutes into the quotient rule: $C'(t) = \\frac{4(2t + 1)^{1/2} - (4t + 8)(2t + 1)^{-1/2}}{2t + 1}$ and simplifies by multiplying the numerator and denominator by $(2t + 1)^{1/2}$ M1\n- Reaches the simplest form: $C'(t) = \\frac{4(2t + 1) - (4t + 8)}{(2t + 1)^{3/2}} = \\frac{4t - 4}{(2t + 1)^{3/2}}$ or $\\frac{4(t - 1)}{(2t + 1)^{3/2}}$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"quotient_rule","label":"Quotient Rule","steps":[{"type":"text","order":0,"title":"Identify quotient rule setup","content":"To find the derivative of $C(t) = \\frac{4t + 8}{\\sqrt{2t + 1}}$, we can use the **Quotient Rule** because the function is written in the form $\\frac{u}{v}$.\n\nThe Quotient Rule states:\n\n$$C'(t) = \\frac{u'v - uv'}{v^2}$$\n\nLet's define our parts:\n- $u = 4t + 8$\n- $v = \\sqrt{2t + 1} = (2t + 1)^{1/2}$","highlights":[{"text":"C(t) = \\frac{4t + 8}{\\sqrt{2t + 1}}","location":"specification"}]},{"type":"text","order":1,"title":"Differentiate u and v","content":"Now, we find the derivatives of $u$ and $v$ with respect to $t$.\n\nFor $u = 4t + 8$:\n$$u' = 4$$\n\nFor $v = (2t + 1)^{1/2}$, we use the **Chain Rule**:\n$$v' = \\frac{1}{2}(2t + 1)^{-1/2} \\times \\frac{d}{dt}(2t + 1)$$\n$$v' = \\frac{1}{2}(2t + 1)^{-1/2} \\times 2 = (2t + 1)^{-1/2}$$\n\nThese individual derivatives are worth marks in the exam markscheme."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u' = \\htmlData{anchor=u-prime-val}{\\color{@sky}{4}}, \\quad v' = \\htmlData{anchor=v-prime-val}{\\color{@pink}{(2t + 1)^{-1/2}}}$"},{"latex":"$$C'(t) = \\frac{\\htmlData{anchor=u-prime-sub}{\\color{@sky}{u'}}v - u\\htmlData{anchor=v-prime-sub}{\\color{@pink}{v'}}}{v^2}$"},{"latex":"$$C'(t) = \\frac{\\color{@sky}{4}\\htmlData{anchor=v-val}{\\color{@amber}{(2t + 1)^{1/2}}} - \\htmlData{anchor=u-val}{\\color{@purple}{(4t + 8)}}\\color{@pink}{(2t + 1)^{-1/2}}}{\\htmlData{anchor=v-sq}{\\color{@teal}{2t + 1}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-prime-val"},{"color":"pink","style":"text-color","anchor":"v-prime-val"},{"color":"amber","style":"underline","anchor":"v-val"},{"color":"purple","style":"underline","anchor":"u-val"},{"color":"teal","style":"box","anchor":"v-sq"}],"connections":[{"to":"u-prime-sub","from":"u-prime-val","color":"sky","label":"u'","style":"solid"},{"to":"v-prime-sub","from":"v-prime-val","color":"pink","label":"v'","style":"solid"}]},"order":2,"title":"Substitute into quotient rule","description":"Substitute $u$, $v$, $u'$, and $v'$ back into the quotient rule formula."},{"type":"text","order":3,"title":"Simplify the fraction","content":"To simplify the expression, we can eliminate the fraction in the numerator by multiplying both the numerator and the denominator by $(2t + 1)^{1/2}$:\n\n$$C'(t) = \\frac{\\left[4(2t + 1)^{1/2} - (4t + 8)(2t + 1)^{-1/2}\\right] \\times (2t + 1)^{1/2}}{(2t + 1) \\times (2t + 1)^{1/2}}$$\n\nApplying this distribution to the numerator:\n- $4(2t + 1)^{1/2} \\times (2t + 1)^{1/2} = 4(2t + 1)$\n- $(4t + 8)(2t + 1)^{-1/2} \\times (2t + 1)^{1/2} = 4t + 8$\n\nAnd for the denominator, using index laws:\n- $(2t + 1)^1 \\times (2t + 1)^{1/2} = (2t + 1)^{3/2}$\n\nThis gives us:\n\n$$C'(t) = \\frac{4(2t + 1) - (4t + 8)}{(2t + 1)^{3/2}}$$"},{"type":"text","order":4,"title":"Write final simplest form","content":"Now expand and collect like terms in the numerator:\n\n$$4(2t + 1) - (4t + 8) = 8t + 4 - 4t - 8$$\n$$= 4t - 4$$\n\nFactoring out a $4$, we can write our final answer in its simplest form:\n\n$$C'(t) = \\frac{4t - 4}{(2t + 1)^{3/2}} = \\frac{4(t - 1)}{(2t + 1)^{3/2}}$$"}]},{"id":"product_rule","label":"Product Rule","steps":[{"type":"text","order":0,"title":"Rewrite as a product","content":"To apply the product rule, we can rewrite the quotient as a product by expressing the square root in the denominator with a negative fractional exponent:\n\n$$C(t) = (4t + 8)(2t + 1)^{-\\frac{1}{2}}$$\n\nNow, we can identify our two functions for the product rule:\n- $$u = 4t + 8$$\n- $$v = (2t + 1)^{-\\frac{1}{2}}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=u-expr}{u = 4t + 8} \\implies \\htmlData{anchor=u-deriv}{\\frac{du}{dt} = 4}$"},{"latex":"$$\\htmlData{anchor=v-expr}{v = (2t + 1)^{-\\frac{1}{2}}} \\implies \\htmlData{anchor=v-deriv}{\\frac{dv}{dt} = -\\frac{1}{2}(2t + 1)^{-\\frac{3}{2}} \\times 2 = -(2t + 1)^{-\\frac{3}{2}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-deriv"},{"color":"pink","style":"text-color","anchor":"v-deriv"}],"connections":[{"to":"u-deriv","from":"u-expr","color":"sky","label":"derivative","style":"solid"},{"to":"v-deriv","from":"v-expr","color":"pink","label":"chain rule","style":"solid"}]},"order":1,"title":"Differentiate each part","description":"We find the derivative of each part. For $v$, we use the chain rule: multiply by the exponent $-\\frac{1}{2}$, subtract 1 from the exponent to get $-\\frac{3}{2}$, and then multiply by the derivative of the inner function, which is $2$."},{"type":"text","order":2,"title":"Apply the product rule","content":"The product rule states that:\n\n$$C'(t) = u \\frac{dv}{dt} + v \\frac{du}{dt}$$\n\nSubstituting our functions and their derivatives:\n\n$$C'(t) = (4t + 8)\\left[-(2t + 1)^{-\\frac{3}{2}}\\right] + (2t + 1)^{-\\frac{1}{2}}(4)$$\n\nRearranging the terms slightly:\n\n$$C'(t) = -(4t + 8)(2t + 1)^{-\\frac{3}{2}} + 4(2t + 1)^{-\\frac{1}{2}}$$"},{"type":"text","order":3,"title":"Factor out common terms","content":"To simplify, we can factor out the common term with the lowest power, which is $(2t + 1)^{-\\frac{3}{2}}$. \n\nRecall that:\n$$(2t + 1)^{-\\frac{1}{2}} = (2t + 1)^{-\\frac{3}{2}}(2t + 1)^1$$\n\nFactoring $(2t + 1)^{-\\frac{3}{2}}$ out of both terms gives:\n\n$$C'(t) = (2t + 1)^{-\\frac{3}{2}} \\left[ -(4t + 8) + 4(2t + 1) \\right]$$"},{"type":"text","order":4,"title":"Simplify and state answer","content":"Now we simplify the terms inside the square brackets:\n\n$$\\begin{aligned} -(4t + 8) + 4(2t + 1) &= -4t - 8 + 8t + 4 \\\\ &= 4t - 4 \\\\ &= 4(t - 1) \\end{aligned}$$\n\nPutting it all together, we can write the negative exponent as a denominator:\n\n$$C'(t) = \\frac{4t - 4}{(2t + 1)^{\\frac{3}{2}}} = \\frac{4(t - 1)}{(2t + 1)^{\\frac{3}{2}}}$$"}]},{"id":"algebraic_decomposition","label":"Algebraic Decomposition","steps":[{"type":"text","order":0,"title":"Decompose the numerator","content":"To differentiate $C(t)$ using algebraic decomposition, we first express the numerator, $4t + 8$, in terms of the linear factor in the denominator, $2t + 1$. \n\nWe can write:\n$$4t + 8 = 2(2t + 1) + 6$$\n\nLet's verify this:\n$$2(2t + 1) + 6 = 4t + 2 + 6 = 4t + 8$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$C(t) = \\frac{\\htmlData{anchor=num}{\\color{@sky}{2(2t + 1) + 6}}}{\\htmlData{anchor=den}{\\color{@amber}{\\sqrt{2t + 1}}}}$"},{"latex":"$$C(t) = \\frac{2(2t + 1)}{\\sqrt{2t + 1}} + \\frac{6}{\\sqrt{2t + 1}}$"},{"latex":"$$C(t) = \\htmlData{anchor=term1}{\\color{@purple}{2(2t + 1)^{1/2}}} + \\htmlData{anchor=term2}{\\color{@teal}{6(2t + 1)^{-1/2}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"num"},{"color":"amber","style":"text-color","anchor":"den"},{"color":"purple","style":"text-color","anchor":"term1"},{"color":"teal","style":"text-color","anchor":"term2"}],"connections":[{"to":"term1","from":"num","color":"sky","label":"split & simplify","style":"solid"}]},"order":1,"title":"Split the fraction","description":"Now, substitute this back into the expression for $C(t)$ and split the fraction into two separate terms, rewriting the square roots as fractional exponents."},{"type":"text","order":2,"title":"Differentiate each term","content":"Next, differentiate each term individually using the chain rule $\\frac{d}{dt}[f(g(t))] = f'(g(t)) \\cdot g'(t)$ where $g(t) = 2t + 1$ (and $g'(t) = 2$):\n\n1. For the first term, $2(2t + 1)^{1/2}$:\n$$\\frac{d}{dt}\\left[2(2t + 1)^{1/2}\\right] = 2 \\cdot \\frac{1}{2}(2t + 1)^{-1/2} \\cdot 2 = 2(2t + 1)^{-1/2}$$\n\n2. For the second term, $6(2t + 1)^{-1/2}$:\n$$\\frac{d}{dt}\\left[6(2t + 1)^{-1/2}\\right] = 6 \\cdot \\left(-\\frac{1}{2}\\right)(2t + 1)^{-3/2} \\cdot 2 = -6(2t + 1)^{-3/2}$$\n\nCombining these, we get:\n$$C'(t) = 2(2t + 1)^{-1/2} - 6(2t + 1)^{-3/2}$$"},{"type":"text","order":3,"title":"Simplify to final form","content":"To write $C'(t)$ in its simplest single-fraction form, factor out the common term with the lowest power, which is $2(2t + 1)^{-3/2}$:\n\n$$\\begin{aligned} C'(t) &= 2(2t + 1)^{-3/2} \\left[ (2t + 1)^{1} - 3 \\right] \\\\ &= 2(2t + 1)^{-3/2} (2t - 2) \\\\ &= 4(t - 1)(2t + 1)^{-3/2} \\end{aligned}$$\n\nExpressing this as a fraction gives the final simplified answer:\n$$C'(t) = \\frac{4(t - 1)}{(2t + 1)^{3/2}}$$"}]}],"generatedAt":"2026-06-09T04:30:32.757Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"90dab143-2e7c-4d9c-87ef-8d2308ae51bc","label":"b","content":"Hence, determine the range of values for the concentration $C$.","markscheme":"- Sets $C'(t) = 0$ to find the stationary point: $4t - 4 = 0 \\implies t = 1$ M1\n- Evaluates the concentration at the stationary point: $C(1) = \\frac{4(1) + 8}{\\sqrt{2(1) + 1}} = \\frac{12}{\\sqrt{3}}$ M1\n- Simplifies the minimum value to $4\\sqrt{3}$ and states the range as $C \\ge 4\\sqrt{3}$ (accept $C \\ge 6.93$ or $y \\ge 4\\sqrt{3}$) A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"calculus","label":"Analytical Calculus Approach","steps":[{"type":"text","order":0,"title":"Define the function","content":"To find the range of the concentration function $C(t)$, we must find its minimum value. We start with the given function for $t > -0.5$:\n\n$$C(t) = \\frac{4t + 8}{\\sqrt{2t + 1}}$$"},{"type":"text","order":1,"title":"Differentiate using Quotient Rule","content":"To find the stationary points, we differentiate $C(t)$ with respect to $t$. We will apply the quotient rule:\n\n$$\\frac{d}{dt}\\left(\\frac{u}{v}\\right) = \\frac{u'v - uv'}{v^2}$$\n\nwhere:\n- $u = 4t + 8 \\implies u' = 4$\n- $v = (2t + 1)^{1/2} \\implies v' = \\frac{1}{2}(2t + 1)^{-1/2}(2) = (2t + 1)^{-1/2}$\n\nSubstituting these into the quotient rule formula gives:\n\n$$C'(t) = \\frac{4\\sqrt{2t + 1} - (4t + 8)(2t + 1)^{-1/2}}{(\\sqrt{2t + 1})^2}$$"},{"type":"text","order":2,"title":"Simplify the derivative","content":"To simplify the numerator of $C'(t)$, we factor out $(2t + 1)^{-1/2}$:\n\n$$C'(t) = \\frac{(2t + 1)^{-1/2} \\left[ 4(2t + 1) - (4t + 8) \\right]}{2t + 1}$$\n\n$$C'(t) = \\frac{8t + 4 - 4t - 8}{(2t + 1)^{3/2}}$$\n\n$$C'(t) = \\frac{4t - 4}{(2t + 1)^{3/2}}$$"},{"type":"text","order":3,"title":"Find the stationary point","content":"To find the stationary point, we set $C'(t) = 0$ to earn the method mark **[M1]**:\n\n$$\\frac{4t - 4}{(2t + 1)^{3/2}} = 0$$\n\nSince the denominator is positive for all $t > -0.5$, we set the numerator to zero:\n\n$$4t - 4 = 0 \\implies t = 1$$"},{"type":"text","order":4,"title":"Find the minimum value","content":"Now, we evaluate the concentration function at the stationary point $t = 1$ to find the minimum value (earning **[M1]**):\n\n$$C(1) = \\frac{4(1) + 8}{\\sqrt{2(1) + 1}} = \\frac{12}{\\sqrt{3}}$$\n\nBy rationalising the denominator:\n\n$$C(1) = \\frac{12\\sqrt{3}}{3} = 4\\sqrt{3} \\approx 6.93$$"},{"type":"text","order":5,"title":"State the range","content":"Since $C(t) \\to \\infty$ as $t \\to \\infty$, the stationary point at $t=1$ is a global minimum. Thus, the concentration is always greater than or equal to this minimum value.\n\nThe range of $C$ is (earning **[A1]**):\n\n$$C \\ge 4\\sqrt{3}$$"}]},{"id":"graphical","label":"Graphical Calculator Approach","steps":[{"type":"text","order":0,"title":"Define the function on your calculator","content":"To find the range of the concentration function, we can use a graphing tool to visualize $C(t)$ for $t > -0.5$.\n\nWe define the function $y = C(x)$ where:\n\n$$y = \\frac{4x + 8}{\\sqrt{2x + 1}}$$\n\nSet your horizontal axis variable to $x$ and set the viewing window to capture the domain $x > -0.5$. A suitable window is $x \\in [-0.5, 5]$ and $y \\in [0, 20]$."},{"type":"text","order":1,"title":"Locate the minimum point","content":"Observing the graph, we can see that as $x$ approaches $-0.5$ from the right, the curve rises steeply towards infinity. As $x$ increases towards positive infinity, the curve also grows larger. This means there is a global minimum value on this interval.\n\nUse the minimum finder tool on your calculator to determine the coordinates of this lowest point."},{"type":"text","order":2,"title":"State the range","content":"From the calculator, the minimum point is located at $(1, 6.9282)$. Let's write the exact value of this minimum to show clear mathematical working:\n\n$$C(1) = \\frac{4(1) + 8}{\\sqrt{2(1) + 1}} = \\frac{12}{\\sqrt{3}}$$\n\nWe can simplify this fraction by rationalising the denominator:\n\n$$C(1) = \\frac{12\\sqrt{3}}{3} = 4\\sqrt{3}$$\n\nSince $4\\sqrt{3} \\approx 6.93$, and the graph has a minimum at this value before rising to infinity on both sides, the range of the concentration $C$ is all values greater than or equal to this minimum:\n\n$$C \\ge 4\\sqrt{3} \\quad \\text{or} \\quad C \\ge 6.93$$"}]}],"generatedAt":"2026-06-09T04:32:37.648Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"7e86b414-ab24-4f77-bc77-70398962ff40:quotient_rule","label":"(a) Quotient Rule","steps":[{"type":"text","order":0,"title":"Identify quotient rule setup","content":"To find the derivative of $C(t) = \\frac{4t + 8}{\\sqrt{2t + 1}}$, we can use the **Quotient Rule** because the function is written in the form $\\frac{u}{v}$.\n\nThe Quotient Rule states:\n\n$$C'(t) = \\frac{u'v - uv'}{v^2}$$\n\nLet's define our parts:\n- $u = 4t + 8$\n- $v = \\sqrt{2t + 1} = (2t + 1)^{1/2}$","highlights":[{"text":"C(t) = \\frac{4t + 8}{\\sqrt{2t + 1}}","location":"specification"}]},{"type":"text","order":1,"title":"Differentiate u and v","content":"Now, we find the derivatives of $u$ and $v$ with respect to $t$.\n\nFor $u = 4t + 8$:\n$$u' = 4$$\n\nFor $v = (2t + 1)^{1/2}$, we use the **Chain Rule**:\n$$v' = \\frac{1}{2}(2t + 1)^{-1/2} \\times \\frac{d}{dt}(2t + 1)$$\n$$v' = \\frac{1}{2}(2t + 1)^{-1/2} \\times 2 = (2t + 1)^{-1/2}$$\n\nThese individual derivatives are worth marks in the exam markscheme."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u' = \\htmlData{anchor=u-prime-val}{\\color{@sky}{4}}, \\quad v' = \\htmlData{anchor=v-prime-val}{\\color{@pink}{(2t + 1)^{-1/2}}}$"},{"latex":"$$C'(t) = \\frac{\\htmlData{anchor=u-prime-sub}{\\color{@sky}{u'}}v - u\\htmlData{anchor=v-prime-sub}{\\color{@pink}{v'}}}{v^2}$"},{"latex":"$$C'(t) = \\frac{\\color{@sky}{4}\\htmlData{anchor=v-val}{\\color{@amber}{(2t + 1)^{1/2}}} - \\htmlData{anchor=u-val}{\\color{@purple}{(4t + 8)}}\\color{@pink}{(2t + 1)^{-1/2}}}{\\htmlData{anchor=v-sq}{\\color{@teal}{2t + 1}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-prime-val"},{"color":"pink","style":"text-color","anchor":"v-prime-val"},{"color":"amber","style":"underline","anchor":"v-val"},{"color":"purple","style":"underline","anchor":"u-val"},{"color":"teal","style":"box","anchor":"v-sq"}],"connections":[{"to":"u-prime-sub","from":"u-prime-val","color":"sky","label":"u'","style":"solid"},{"to":"v-prime-sub","from":"v-prime-val","color":"pink","label":"v'","style":"solid"}]},"order":2,"title":"Substitute into quotient rule","description":"Substitute $u$, $v$, $u'$, and $v'$ back into the quotient rule formula."},{"type":"text","order":3,"title":"Simplify the fraction","content":"To simplify the expression, we can eliminate the fraction in the numerator by multiplying both the numerator and the denominator by $(2t + 1)^{1/2}$:\n\n$$C'(t) = \\frac{\\left[4(2t + 1)^{1/2} - (4t + 8)(2t + 1)^{-1/2}\\right] \\times (2t + 1)^{1/2}}{(2t + 1) \\times (2t + 1)^{1/2}}$$\n\nApplying this distribution to the numerator:\n- $4(2t + 1)^{1/2} \\times (2t + 1)^{1/2} = 4(2t + 1)$\n- $(4t + 8)(2t + 1)^{-1/2} \\times (2t + 1)^{1/2} = 4t + 8$\n\nAnd for the denominator, using index laws:\n- $(2t + 1)^1 \\times (2t + 1)^{1/2} = (2t + 1)^{3/2}$\n\nThis gives us:\n\n$$C'(t) = \\frac{4(2t + 1) - (4t + 8)}{(2t + 1)^{3/2}}$$"},{"type":"text","order":4,"title":"Write final simplest form","content":"Now expand and collect like terms in the numerator:\n\n$$4(2t + 1) - (4t + 8) = 8t + 4 - 4t - 8$$\n$$= 4t - 4$$\n\nFactoring out a $4$, we can write our final answer in its simplest form:\n\n$$C'(t) = \\frac{4t - 4}{(2t + 1)^{3/2}} = \\frac{4(t - 1)}{(2t + 1)^{3/2}}$$"}]},{"id":"7e86b414-ab24-4f77-bc77-70398962ff40:product_rule","label":"(a) Product Rule","steps":[{"type":"text","order":0,"title":"Rewrite as a product","content":"To apply the product rule, we can rewrite the quotient as a product by expressing the square root in the denominator with a negative fractional exponent:\n\n$$C(t) = (4t + 8)(2t + 1)^{-\\frac{1}{2}}$$\n\nNow, we can identify our two functions for the product rule:\n- $$u = 4t + 8$$\n- $$v = (2t + 1)^{-\\frac{1}{2}}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=u-expr}{u = 4t + 8} \\implies \\htmlData{anchor=u-deriv}{\\frac{du}{dt} = 4}$"},{"latex":"$$\\htmlData{anchor=v-expr}{v = (2t + 1)^{-\\frac{1}{2}}} \\implies \\htmlData{anchor=v-deriv}{\\frac{dv}{dt} = -\\frac{1}{2}(2t + 1)^{-\\frac{3}{2}} \\times 2 = -(2t + 1)^{-\\frac{3}{2}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-deriv"},{"color":"pink","style":"text-color","anchor":"v-deriv"}],"connections":[{"to":"u-deriv","from":"u-expr","color":"sky","label":"derivative","style":"solid"},{"to":"v-deriv","from":"v-expr","color":"pink","label":"chain rule","style":"solid"}]},"order":1,"title":"Differentiate each part","description":"We find the derivative of each part. For $v$, we use the chain rule: multiply by the exponent $-\\frac{1}{2}$, subtract 1 from the exponent to get $-\\frac{3}{2}$, and then multiply by the derivative of the inner function, which is $2$."},{"type":"text","order":2,"title":"Apply the product rule","content":"The product rule states that:\n\n$$C'(t) = u \\frac{dv}{dt} + v \\frac{du}{dt}$$\n\nSubstituting our functions and their derivatives:\n\n$$C'(t) = (4t + 8)\\left[-(2t + 1)^{-\\frac{3}{2}}\\right] + (2t + 1)^{-\\frac{1}{2}}(4)$$\n\nRearranging the terms slightly:\n\n$$C'(t) = -(4t + 8)(2t + 1)^{-\\frac{3}{2}} + 4(2t + 1)^{-\\frac{1}{2}}$$"},{"type":"text","order":3,"title":"Factor out common terms","content":"To simplify, we can factor out the common term with the lowest power, which is $(2t + 1)^{-\\frac{3}{2}}$. \n\nRecall that:\n$$(2t + 1)^{-\\frac{1}{2}} = (2t + 1)^{-\\frac{3}{2}}(2t + 1)^1$$\n\nFactoring $(2t + 1)^{-\\frac{3}{2}}$ out of both terms gives:\n\n$$C'(t) = (2t + 1)^{-\\frac{3}{2}} \\left[ -(4t + 8) + 4(2t + 1) \\right]$$"},{"type":"text","order":4,"title":"Simplify and state answer","content":"Now we simplify the terms inside the square brackets:\n\n$$\\begin{aligned} -(4t + 8) + 4(2t + 1) &= -4t - 8 + 8t + 4 \\\\ &= 4t - 4 \\\\ &= 4(t - 1) \\end{aligned}$$\n\nPutting it all together, we can write the negative exponent as a denominator:\n\n$$C'(t) = \\frac{4t - 4}{(2t + 1)^{\\frac{3}{2}}} = \\frac{4(t - 1)}{(2t + 1)^{\\frac{3}{2}}}$$"}]},{"id":"7e86b414-ab24-4f77-bc77-70398962ff40:algebraic_decomposition","label":"(a) Algebraic Decomposition","steps":[{"type":"text","order":0,"title":"Decompose the numerator","content":"To differentiate $C(t)$ using algebraic decomposition, we first express the numerator, $4t + 8$, in terms of the linear factor in the denominator, $2t + 1$. \n\nWe can write:\n$$4t + 8 = 2(2t + 1) + 6$$\n\nLet's verify this:\n$$2(2t + 1) + 6 = 4t + 2 + 6 = 4t + 8$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$C(t) = \\frac{\\htmlData{anchor=num}{\\color{@sky}{2(2t + 1) + 6}}}{\\htmlData{anchor=den}{\\color{@amber}{\\sqrt{2t + 1}}}}$"},{"latex":"$$C(t) = \\frac{2(2t + 1)}{\\sqrt{2t + 1}} + \\frac{6}{\\sqrt{2t + 1}}$"},{"latex":"$$C(t) = \\htmlData{anchor=term1}{\\color{@purple}{2(2t + 1)^{1/2}}} + \\htmlData{anchor=term2}{\\color{@teal}{6(2t + 1)^{-1/2}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"num"},{"color":"amber","style":"text-color","anchor":"den"},{"color":"purple","style":"text-color","anchor":"term1"},{"color":"teal","style":"text-color","anchor":"term2"}],"connections":[{"to":"term1","from":"num","color":"sky","label":"split & simplify","style":"solid"}]},"order":1,"title":"Split the fraction","description":"Now, substitute this back into the expression for $C(t)$ and split the fraction into two separate terms, rewriting the square roots as fractional exponents."},{"type":"text","order":2,"title":"Differentiate each term","content":"Next, differentiate each term individually using the chain rule $\\frac{d}{dt}[f(g(t))] = f'(g(t)) \\cdot g'(t)$ where $g(t) = 2t + 1$ (and $g'(t) = 2$):\n\n1. For the first term, $2(2t + 1)^{1/2}$:\n$$\\frac{d}{dt}\\left[2(2t + 1)^{1/2}\\right] = 2 \\cdot \\frac{1}{2}(2t + 1)^{-1/2} \\cdot 2 = 2(2t + 1)^{-1/2}$$\n\n2. For the second term, $6(2t + 1)^{-1/2}$:\n$$\\frac{d}{dt}\\left[6(2t + 1)^{-1/2}\\right] = 6 \\cdot \\left(-\\frac{1}{2}\\right)(2t + 1)^{-3/2} \\cdot 2 = -6(2t + 1)^{-3/2}$$\n\nCombining these, we get:\n$$C'(t) = 2(2t + 1)^{-1/2} - 6(2t + 1)^{-3/2}$$"},{"type":"text","order":3,"title":"Simplify to final form","content":"To write $C'(t)$ in its simplest single-fraction form, factor out the common term with the lowest power, which is $2(2t + 1)^{-3/2}$:\n\n$$\\begin{aligned} C'(t) &= 2(2t + 1)^{-3/2} \\left[ (2t + 1)^{1} - 3 \\right] \\\\ &= 2(2t + 1)^{-3/2} (2t - 2) \\\\ &= 4(t - 1)(2t + 1)^{-3/2} \\end{aligned}$$\n\nExpressing this as a fraction gives the final simplified answer:\n$$C'(t) = \\frac{4(t - 1)}{(2t + 1)^{3/2}}$$"}]},{"id":"90dab143-2e7c-4d9c-87ef-8d2308ae51bc:calculus","label":"(b) Analytical Calculus Approach","steps":[{"type":"text","order":0,"title":"Define the function","content":"To find the range of the concentration function $C(t)$, we must find its minimum value. We start with the given function for $t > -0.5$:\n\n$$C(t) = \\frac{4t + 8}{\\sqrt{2t + 1}}$$"},{"type":"text","order":1,"title":"Differentiate using Quotient Rule","content":"To find the stationary points, we differentiate $C(t)$ with respect to $t$. We will apply the quotient rule:\n\n$$\\frac{d}{dt}\\left(\\frac{u}{v}\\right) = \\frac{u'v - uv'}{v^2}$$\n\nwhere:\n- $u = 4t + 8 \\implies u' = 4$\n- $v = (2t + 1)^{1/2} \\implies v' = \\frac{1}{2}(2t + 1)^{-1/2}(2) = (2t + 1)^{-1/2}$\n\nSubstituting these into the quotient rule formula gives:\n\n$$C'(t) = \\frac{4\\sqrt{2t + 1} - (4t + 8)(2t + 1)^{-1/2}}{(\\sqrt{2t + 1})^2}$$"},{"type":"text","order":2,"title":"Simplify the derivative","content":"To simplify the numerator of $C'(t)$, we factor out $(2t + 1)^{-1/2}$:\n\n$$C'(t) = \\frac{(2t + 1)^{-1/2} \\left[ 4(2t + 1) - (4t + 8) \\right]}{2t + 1}$$\n\n$$C'(t) = \\frac{8t + 4 - 4t - 8}{(2t + 1)^{3/2}}$$\n\n$$C'(t) = \\frac{4t - 4}{(2t + 1)^{3/2}}$$"},{"type":"text","order":3,"title":"Find the stationary point","content":"To find the stationary point, we set $C'(t) = 0$ to earn the method mark **[M1]**:\n\n$$\\frac{4t - 4}{(2t + 1)^{3/2}} = 0$$\n\nSince the denominator is positive for all $t > -0.5$, we set the numerator to zero:\n\n$$4t - 4 = 0 \\implies t = 1$$"},{"type":"text","order":4,"title":"Find the minimum value","content":"Now, we evaluate the concentration function at the stationary point $t = 1$ to find the minimum value (earning **[M1]**):\n\n$$C(1) = \\frac{4(1) + 8}{\\sqrt{2(1) + 1}} = \\frac{12}{\\sqrt{3}}$$\n\nBy rationalising the denominator:\n\n$$C(1) = \\frac{12\\sqrt{3}}{3} = 4\\sqrt{3} \\approx 6.93$$"},{"type":"text","order":5,"title":"State the range","content":"Since $C(t) \\to \\infty$ as $t \\to \\infty$, the stationary point at $t=1$ is a global minimum. Thus, the concentration is always greater than or equal to this minimum value.\n\nThe range of $C$ is (earning **[A1]**):\n\n$$C \\ge 4\\sqrt{3}$$"}]},{"id":"90dab143-2e7c-4d9c-87ef-8d2308ae51bc:graphical","label":"(b) Graphical Calculator Approach","steps":[{"type":"text","order":0,"title":"Define the function on your calculator","content":"To find the range of the concentration function, we can use a graphing tool to visualize $C(t)$ for $t > -0.5$.\n\nWe define the function $y = C(x)$ where:\n\n$$y = \\frac{4x + 8}{\\sqrt{2x + 1}}$$\n\nSet your horizontal axis variable to $x$ and set the viewing window to capture the domain $x > -0.5$. A suitable window is $x \\in [-0.5, 5]$ and $y \\in [0, 20]$."},{"type":"text","order":1,"title":"Locate the minimum point","content":"Observing the graph, we can see that as $x$ approaches $-0.5$ from the right, the curve rises steeply towards infinity. As $x$ increases towards positive infinity, the curve also grows larger. This means there is a global minimum value on this interval.\n\nUse the minimum finder tool on your calculator to determine the coordinates of this lowest point."},{"type":"text","order":2,"title":"State the range","content":"From the calculator, the minimum point is located at $(1, 6.9282)$. Let's write the exact value of this minimum to show clear mathematical working:\n\n$$C(1) = \\frac{4(1) + 8}{\\sqrt{2(1) + 1}} = \\frac{12}{\\sqrt{3}}$$\n\nWe can simplify this fraction by rationalising the denominator:\n\n$$C(1) = \\frac{12\\sqrt{3}}{3} = 4\\sqrt{3}$$\n\nSince $4\\sqrt{3} \\approx 6.93$, and the graph has a minimum at this value before rising to infinity on both sides, the range of the concentration $C$ is all values greater than or equal to this minimum:\n\n$$C \\ge 4\\sqrt{3} \\quad \\text{or} \\quad C \\ge 6.93$$"}]}],"generatedAt":"2026-06-09T04:32:52.922Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ocr_mei","summaries":[{"id":"43670804-b308-4c43-b1b0-15ba09240572","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"233cef29-b34e-452d-b329-eeb2f8d1af50","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"d7542af0-c3ba-44ed-81cd-ab6ab6f8a30a","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"dcc84617-b6eb-4564-b8e5-ed5142a90f67","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"0720e9a2-40c4-4527-8636-f1396bc14a09","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"1fe24200-387c-4d5a-aadd-796df3778a45","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"9548d2cd-09f8-45c9-8239-a49aa6d6bb6d","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"e90e6e7d-87c9-419a-b761-350ef207d2f5","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"62776d45-a6ee-448a-bde0-84f38af196c4","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"78948c35-3ae0-4e85-af96-a4ab24b2a905","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"fc58532d-2057-4310-9ca0-ac9832b1dbe5","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"1b1db3e5-c4d6-446a-ba82-54b917bb77f8","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"619f22d8-6582-4a8e-a8c8-fca3c4302e31","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"02a4484d-6465-47f9-8569-d80bc542a95c","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"0de699ee-92d8-42c4-a6ff-32b84bb6e495","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"137c633d-888b-442a-b962-213f13a99168","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1dd6e7ed-234f-4728-8cdb-9ab46caf8736","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":false},{"id":"362274ef-ce53-4c09-a816-30ac6ec89d5d","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"39e689e7-2e31-4a3a-9d70-526672dcf8d9","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"56a69fac-7065-42e6-b30b-a21da9d2291a","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"6b1223c3-3eb3-46d9-b189-b43dfd983e74","difficultyBand":"hard","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"b3657bab-3488-4ffa-af71-3222ec757ec3","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"ccae4139-d718-4443-8d1b-9e52e56a46fe","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"d5404ddf-d8b4-40da-8504-5e685b72f033","difficultyBand":"medium","calculatorUse":null,"totalMarks":15,"questionType":"LA","isCalculation":true},{"id":"d78e5e76-d516-4aec-8878-4041448d7b0c","difficultyBand":"hard","calculatorUse":null,"totalMarks":16,"questionType":"LA","isCalculation":true},{"id":"e963c301-9c90-428f-8a7e-ce3d26f8289c","difficultyBand":"hard","calculatorUse":null,"totalMarks":13,"questionType":"LA","isCalculation":true},{"id":"ec691163-01bf-44ff-9d8e-59875ab7ab0a","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"f741e45a-3beb-4b3f-8712-14a39d812f97","difficultyBand":"hard","calculatorUse":null,"totalMarks":13,"questionType":"LA","isCalculation":true},{"id":"f7c42d77-ed0f-4ad7-8c97-750efd22a0ad","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true}],"title":"1.9.13 Quotient rule (A-level only)","topicId":"2af73925-f8b7-4780-a294-32d3d331b72e","topicSidebar":"$L3c","topicSlug":"quotient-rule","topicTitle":"1.9.13 Quotient rule (A-level only)"}]}],"$L3d","$L3e"]}]

1.9.13 Quotient rule (A-level only)

Question 16

1.9.13 Quotient rule (A-level only) Questions

1.9.13 Quotient rule (A-level only)

Question 16

1.9.13 Quotient rule (A-level only) Questions