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rule","style":"solid"}]},"order":1,"title":"Differentiate each term","description":"We can now differentiate each term individually. The derivative of $x$ is simply $1$. For the second term, we apply the chain rule: multiply by the power, decrease the power by 1, and then multiply by the derivative of the inside bracket."},{"type":"text","order":2,"title":"Apply the chain rule","content":"The derivative of the inner function $(x^2 + 6)$ is:\n\n$$\\frac{d}{dx}(x^2 + 6) = 2x$$\n\nSubstituting this back into our expression for the derivative:\n\n$$\\frac{dy}{dx} = 1 - (-1)(x^2 + 6)^{-2} \\times (2x)$$\n\nThis step earns you the first mark (**M1**) for applying the chain rule correctly."},{"type":"text","order":3,"title":"Simplify to final form","content":"Now we simplify the signs and multiply the terms together:\n\n$$-(-1) \\times 2x = +2x$$\n\nSo, the fully simplified derivative is:\n\n$$\\frac{dy}{dx} = 1 + 2x(x^2 + 6)^{-2}$$\n\nThis final form earns you the accuracy mark (**A1**)."}]},{"id":"quotient_rule_term","label":"Using Quotient Rule on the Fraction","steps":[{"type":"text","order":0,"title":"Break down the function","content":"We need to differentiate the function with respect to $x$:\n\n$$y = x - \\frac{1}{x^2 + 6}$$\n\nUsing the linearity of differentiation, we can differentiate each term separately:\n\n$$\\frac{dy}{dx} = \\frac{d}{dx}(x) - \\frac{d}{dx}\\left(\\frac{1}{x^2 + 6}\\right)$$. "},{"type":"text","order":1,"title":"Differentiate the first term","content":"The derivative of the first term, $x$, with respect to $x$ is straightforward:\n\n$$\\frac{d}{dx}(x) = 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } y_2 = \\frac{\\htmlData{anchor=u-term}{\\color{@sky}{1}}}{\\htmlData{anchor=v-term}{\\color{@amber}{x^2 + 6}}}$"},{"latex":"$$\\text{Here, } \\htmlData{anchor=u-val}{\\color{@sky}{u = 1}} \\implies \\htmlData{anchor=du-val}{\\color{@purple}{u' = 0}}$"},{"latex":"$$\\text{and } \\htmlData{anchor=v-val}{\\color{@amber}{v = x^2 + 6}} \\implies \\htmlData{anchor=dv-val}{\\color{@teal}{v' = 2x}}$"},{"latex":"$$\\frac{d}{dx}\\left(\\frac{u}{v}\\right) = \\frac{\\htmlData{anchor=du-sub}{\\color{@purple}{u'}} \\htmlData{anchor=v-sub}{\\color{@amber}{v}} - \\htmlData{anchor=u-sub}{\\color{@sky}{u}} \\htmlData{anchor=dv-sub}{\\color{@teal}{v'}}}{v^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"v-term"}],"connections":[{"to":"u-val","from":"u-term","color":"sky","label":"numerator","style":"solid"},{"to":"v-val","from":"v-term","color":"amber","label":"denominator","style":"solid"}]},"order":2,"title":"Apply the quotient rule","description":"To differentiate the fraction term $\\frac{1}{x^2 + 6}$, we use the Quotient Rule: $\\frac{d}{dx}\\left(\\frac{u}{v}\\right) = \\frac{u'v - uv'}{v^2}$."},{"type":"text","order":3,"title":"Calculate the fraction's derivative","content":"Now we substitute our functions and their derivatives into the quotient rule formula:\n\n$$\\frac{d}{dx}\\left(\\frac{1}{x^2 + 6}\\right) = \\frac{(0)(x^2 + 6) - (1)(2x)}{(x^2 + 6)^2}$$\n\n$$= \\frac{-2x}{(x^2 + 6)^2}$$\n\nThis would earn you the first method mark **M1** for applying a valid differentiation method to the terms."},{"type":"text","order":4,"title":"Combine the results","content":"Now we combine the derivative of both parts:\n\n$$\\frac{dy}{dx} = 1 - \\left(\\frac{-2x}{(x^2 + 6)^2}\\right)$$\n\n$$= 1 + \\frac{2x}{(x^2 + 6)^2}$$\n\nUsing negative indices, we can write the final answer as:\n\n$$1 + 2x(x^2 + 6)^{-2}$$\n\nThis gives you the final accuracy mark **A1**."}]},{"id":"single_fraction_quotient","label":"Combining and Using Quotient Rule","steps":[{"type":"text","order":0,"title":"Combine into one fraction","content":"To begin with the **Combining and Using Quotient Rule** method, we first express our function as a single algebraic fraction. We can do this by finding a common denominator:\n\n$$y = x - \\frac{1}{x^2 + 6}$$\n\n$$y = \\frac{x(x^2 + 6)}{x^2 + 6} - \\frac{1}{x^2 + 6}$$\n\n$$y = \\frac{x^3 + 6x - 1}{x^2 + 6}$$. "},{"type":"text","order":1,"title":"Identify u and v","content":"Now we can apply the quotient rule, which is found in your formulas for differentiation. If $y = \\frac{u}{v}$, then:\n\n$$\\frac{dy}{dx} = \\frac{u'v - uv'}{v^2}$$\n\nLet's identify our numerator $u$ and denominator $v$ and differentiate them both with respect to $x$:\n\n- $u = x^3 + 6x - 1 \\implies u' = 3x^2 + 6$\n- $v = x^2 + 6 \\implies v' = 2x$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{\\htmlData{anchor=u-prime-val}{\\color{@sky}{u'}} \\htmlData{anchor=v-val}{\\color{@amber}{v}} - \\htmlData{anchor=u-val}{\\color{@purple}{u}} \\htmlData{anchor=v-prime-val}{\\color{@teal}{v'}}}{\\left(\\htmlData{anchor=v-denom}{\\color{@amber}{v}}\\right)^2}$$"},{"latex":"$$$\\frac{\\htmlData{anchor=u-prime-sub}{\\color{@sky}{(3x^2 + 6)}}\\htmlData{anchor=v-sub}{\\color{@amber}{(x^2 + 6)}} - \\htmlData{anchor=u-sub}{\\color{@purple}{(x^3 + 6x - 1)}}\\htmlData{anchor=v-prime-sub}{\\color{@teal}{(2x)}}}{\\left(\\htmlData{anchor=v-denom-sub}{\\color{@amber}{x^2 + 6}}\\right)^2}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-prime-sub"},{"color":"amber","style":"box","anchor":"v-sub"},{"color":"purple","style":"box","anchor":"u-sub"},{"color":"teal","style":"box","anchor":"v-prime-sub"}],"connections":[{"to":"u-prime-sub","from":"u-prime-val","color":"sky","label":"u'","style":"solid"},{"to":"v-sub","from":"v-val","color":"amber","label":"v","style":"solid"},{"to":"u-sub","from":"u-val","color":"purple","label":"u","style":"solid"},{"to":"v-prime-sub","from":"v-prime-val","color":"teal","label":"v'","style":"solid"}]},"order":2,"title":"Substitute into quotient rule","description":"Substitute our expressions for $u, u', v,$ and $v'$ directly into the quotient rule formula."},{"type":"text","order":3,"title":"Simplify the numerator","content":"Now we expand and simplify the terms in the numerator:\n\n1. Expand the first term:\n\n$$(3x^2 + 6)(x^2 + 6) = 3x^4 + 18x^2 + 6x^2 + 36 = 3x^4 + 24x^2 + 36$$\n\n2. Expand the second term:\n\n$$(x^3 + 6x - 1)(2x) = 2x^4 + 12x^2 - 2x$$\n\n3. Subtract them:\n\n$$\\text{Numerator} = (3x^4 + 24x^2 + 36) - (2x^4 + 12x^2 - 2x)$$\n\n$$= 3x^4 - 2x^4 + 24x^2 - 12x^2 + 2x + 36$$\n\n$$= x^4 + 12x^2 + 2x + 36$$\n\nPutting it all back together over the denominator, we get:\n\n$$\\frac{dy}{dx} = \\frac{x^4 + 12x^2 + 2x + 36}{(x^2 + 6)^2}$$"},{"type":"text","order":4,"title":"Relate to alternative form","content":"The answer we obtained is fully correct! We can also show it is equivalent to the expression in the markscheme, $1 + 2x(x^2 + 6)^{-2}$, by splitting our fraction:\n\n$$\\frac{x^4 + 12x^2 + 2x + 36}{(x^2 + 6)^2} = \\frac{(x^4 + 12x^2 + 36) + 2x}{(x^2 + 6)^2}$$\n\n$$= \\frac{(x^2 + 6)^2 + 2x}{(x^2 + 6)^2}$$\n\n$$= \\frac{(x^2 + 6)^2}{(x^2 + 6)^2} + \\frac{2x}{(x^2 + 6)^2}$$\n\n$$= 1 + \\frac{2x}{(x^2 + 6)^2}$$\n\n$$= 1 + 2x(x^2 + 6)^{-2}$$\n\nBoth forms are completely valid, but combining first into a single fraction is a very robust way to avoid mistakes with negative power indices!"}]}],"generatedAt":"2026-06-09T09:13:13.302Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"e9b524cc-b698-4e5b-ab9d-37ac24370b72","label":"b","content":"$$2 + e^{0.5x^2}$","markscheme":"- Differentiates $e^{0.5x^2}$ to $x e^{0.5x^2}$ M1\n- Final answer $x e^{0.5x^2}$ A1","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Chain rule applied to $e^{0.5x^2}$, resulting in $f'(x)e^{f(x)}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct final derivative expression.","code":"A"}],"finalAnswers":["$$x e^{0.5x^2}$"],"generatedAt":"2026-05-06T14:04:00.767Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_chain_rule","label":"Direct Chain Rule","steps":[{"type":"text","order":0,"title":"Analyze the function","content":"We need to find the derivative of the expression with respect to $x$:\n\n$$\\frac{d}{dx}\\left(2 + e^{0.5x^2}\\right)$$\n\nThis expression is composed of two parts: a constant term, $2$, and an exponential term, $e^{0.5x^2}$. We can differentiate each term individually."},{"type":"text","order":1,"title":"Differentiate the constant","content":"The derivative of any constant term with respect to $x$ is zero:\n\n$$\\frac{d}{dx}(2) = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"For the exponential term, we first identify the exponent:"},{"latex":"$$e^{\\htmlData{anchor=exponent}{\\color{@sky}{0.5x^2}}}$"},{"text":"We then differentiate this exponent with respect to $x$:"},{"latex":"$$$\\frac{d}{dx}(\\htmlData{anchor=exponent-sub}{\\color{@sky}{0.5x^2}}) = \\htmlData{anchor=deriv-exponent}{\\color{@amber}{x}}$$"},{"text":"Finally, we multiply the derivative of the exponent by the original exponential term:"},{"latex":"$$$\\frac{d}{dx}\\left(e^{\\color{@sky}{0.5x^2}}\\right) = \\htmlData{anchor=result-deriv}{\\color{@amber}{x}} e^{\\color{@sky}{0.5x^2}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"amber","style":"text-color","anchor":"deriv-exponent"}],"connections":[{"to":"exponent-sub","from":"exponent","color":"sky","label":null,"style":"solid"},{"to":"result-deriv","from":"deriv-exponent","color":"amber","label":"derivative of exponent","style":"solid"}]},"order":2,"title":"Apply direct chain rule","description":"Using the chain rule, $\\frac{d}{dx}(e^{u}) = u' e^{u}$, we find the derivative of the inner exponent function and multiply it by the original exponential expression."},{"type":"text","order":3,"title":"State the final answer","content":"Combining the derivatives of both terms gives the final result:\n\n$$\\frac{d}{dx}\\left(2 + e^{0.5x^2}\\right) = 0 + x e^{0.5x^2} = x e^{0.5x^2}$$\n\nThis earns the **M1** mark for correctly differentiating the exponential part and the **A1** mark for the correct final simplified expression."}]},{"id":"formal_substitution","label":"Formal u-Substitution","steps":[{"type":"text","order":0,"title":"Define the target function","content":"Let our function be defined as:\n\n$$y = 2 + e^{0.5x^2}$$\n\nWe want to find the derivative $\\frac{dy}{dx}$. \n\nSince this function contains a constant and a composite function, we can differentiate each term individually. The constant term $2$ will differentiate to $0$. To find the derivative of the exponential term $e^{0.5x^2}$, we will use the **Formal $u$-Substitution** method."},{"type":"hyper_latex","block":{"nodes":[{"text":"We define a new variable for the exponent:"},{"latex":"$$u = \\htmlData{anchor=exponent}{\\color{@sky}{0.5x^2}}$"},{"text":"This allows us to write the function $y$ in terms of $u$:"},{"latex":"$$y = 2 + e^{\\htmlData{anchor=u-var}{\\color{@sky}{u}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"sky","style":"text-color","anchor":"u-var"}],"connections":[{"to":"u-var","from":"exponent","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We identify the inner function as the exponent of the natural exponential function and substitute it with a new variable $u$."},{"type":"text","order":2,"title":"Calculate the individual derivatives","content":"Next, we need to find the derivative of $u$ with respect to $x$, and the derivative of $y$ with respect to $u$.\n\n1. Differentiate $u = 0.5x^2$ with respect to $x$ using the power rule:\n $$\\frac{du}{dx} = 0.5 \\cdot 2x = x$$\n\n2. Differentiate $y = 2 + e^u$ with respect to $u$:\n $$\\frac{dy}{du} = e^u$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Now we apply the formal chain rule formula:"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dy-du}{\\color{@teal}{\\frac{dy}{du}}} \\cdot \\htmlData{anchor=du-dx}{\\color{@purple}{\\frac{du}{dx}}}$"},{"text":"Substituting our expressions into this formula:"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dy-du-val}{\\color{@teal}{e^u}} \\cdot \\htmlData{anchor=du-dx-val}{\\color{@purple}{x}}$"}],"highlights":[{"color":"teal","style":"text-color","anchor":"dy-du"},{"color":"purple","style":"text-color","anchor":"du-dx"},{"color":"teal","style":"text-color","anchor":"dy-du-val"},{"color":"purple","style":"text-color","anchor":"du-dx-val"}],"connections":[{"to":"dy-du","from":"dy-du-val","color":"teal","label":"substitute dy/du","style":"solid"},{"to":"du-dx","from":"du-dx-val","color":"purple","label":"substitute du/dx","style":"solid"}]},"order":3,"title":"Apply the Chain Rule","description":"We substitute our derivatives into the Chain Rule formula to find $\\frac{dy}{dx}$ in terms of $x$ and $u$."},{"type":"text","order":4,"title":"Substitute back for x","content":"Finally, we substitute our original expression $u = 0.5x^2$ back into the derivative:\n\n$$\\frac{dy}{dx} = x e^{0.5x^2}$$\n\nThis earns both marks: \n- **M1** for correctly differentiating the exponential term using the chain rule.\n- **A1** for the final simplified answer."}]}],"generatedAt":"2026-06-09T09:15:36.612Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"a0baa34c-5846-47cd-9b63-35d326381dfd:chain_rule","label":"(a) Using Chain Rule and Index Laws","steps":[{"type":"text","order":0,"title":"Rewrite using index laws","content":"To make this expression easier to differentiate using the chain rule, we can rewrite the fraction using a negative exponent.\n\nRecall that:\n\n$$\\frac{1}{f(x)} = [f(x)]^{-1}$$\n\nApplying this to our expression gives:\n\n$$y = x - (x^2 + 6)^{-1}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=term-x}{\\color{@sky}{x}} - \\htmlData{anchor=term-chain}{\\color{@amber}{(x^2 + 6)^{-1}}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=diff-x}{\\color{@sky}{1}} - \\htmlData{anchor=diff-chain}{\\color{@amber}{(-1)(x^2 + 6)^{-2} \\times \\frac{d}{dx}(x^2 + 6)}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term-x"},{"color":"amber","style":"text-color","anchor":"term-chain"}],"connections":[{"to":"diff-x","from":"term-x","color":"sky","label":null,"style":"solid"},{"to":"diff-chain","from":"term-chain","color":"amber","label":"chain rule","style":"solid"}]},"order":1,"title":"Differentiate each term","description":"We can now differentiate each term individually. The derivative of $x$ is simply $1$. For the second term, we apply the chain rule: multiply by the power, decrease the power by 1, and then multiply by the derivative of the inside bracket."},{"type":"text","order":2,"title":"Apply the chain rule","content":"The derivative of the inner function $(x^2 + 6)$ is:\n\n$$\\frac{d}{dx}(x^2 + 6) = 2x$$\n\nSubstituting this back into our expression for the derivative:\n\n$$\\frac{dy}{dx} = 1 - (-1)(x^2 + 6)^{-2} \\times (2x)$$\n\nThis step earns you the first mark (**M1**) for applying the chain rule correctly."},{"type":"text","order":3,"title":"Simplify to final form","content":"Now we simplify the signs and multiply the terms together:\n\n$$-(-1) \\times 2x = +2x$$\n\nSo, the fully simplified derivative is:\n\n$$\\frac{dy}{dx} = 1 + 2x(x^2 + 6)^{-2}$$\n\nThis final form earns you the accuracy mark (**A1**)."}]},{"id":"a0baa34c-5846-47cd-9b63-35d326381dfd:quotient_rule_term","label":"(a) Using Quotient Rule on the Fraction","steps":[{"type":"text","order":0,"title":"Break down the function","content":"We need to differentiate the function with respect to $x$:\n\n$$y = x - \\frac{1}{x^2 + 6}$$\n\nUsing the linearity of differentiation, we can differentiate each term separately:\n\n$$\\frac{dy}{dx} = \\frac{d}{dx}(x) - \\frac{d}{dx}\\left(\\frac{1}{x^2 + 6}\\right)$$. "},{"type":"text","order":1,"title":"Differentiate the first term","content":"The derivative of the first term, $x$, with respect to $x$ is straightforward:\n\n$$\\frac{d}{dx}(x) = 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } y_2 = \\frac{\\htmlData{anchor=u-term}{\\color{@sky}{1}}}{\\htmlData{anchor=v-term}{\\color{@amber}{x^2 + 6}}}$"},{"latex":"$$\\text{Here, } \\htmlData{anchor=u-val}{\\color{@sky}{u = 1}} \\implies \\htmlData{anchor=du-val}{\\color{@purple}{u' = 0}}$"},{"latex":"$$\\text{and } \\htmlData{anchor=v-val}{\\color{@amber}{v = x^2 + 6}} \\implies \\htmlData{anchor=dv-val}{\\color{@teal}{v' = 2x}}$"},{"latex":"$$\\frac{d}{dx}\\left(\\frac{u}{v}\\right) = \\frac{\\htmlData{anchor=du-sub}{\\color{@purple}{u'}} \\htmlData{anchor=v-sub}{\\color{@amber}{v}} - \\htmlData{anchor=u-sub}{\\color{@sky}{u}} \\htmlData{anchor=dv-sub}{\\color{@teal}{v'}}}{v^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"v-term"}],"connections":[{"to":"u-val","from":"u-term","color":"sky","label":"numerator","style":"solid"},{"to":"v-val","from":"v-term","color":"amber","label":"denominator","style":"solid"}]},"order":2,"title":"Apply the quotient rule","description":"To differentiate the fraction term $\\frac{1}{x^2 + 6}$, we use the Quotient Rule: $\\frac{d}{dx}\\left(\\frac{u}{v}\\right) = \\frac{u'v - uv'}{v^2}$."},{"type":"text","order":3,"title":"Calculate the fraction's derivative","content":"Now we substitute our functions and their derivatives into the quotient rule formula:\n\n$$\\frac{d}{dx}\\left(\\frac{1}{x^2 + 6}\\right) = \\frac{(0)(x^2 + 6) - (1)(2x)}{(x^2 + 6)^2}$$\n\n$$= \\frac{-2x}{(x^2 + 6)^2}$$\n\nThis would earn you the first method mark **M1** for applying a valid differentiation method to the terms."},{"type":"text","order":4,"title":"Combine the results","content":"Now we combine the derivative of both parts:\n\n$$\\frac{dy}{dx} = 1 - \\left(\\frac{-2x}{(x^2 + 6)^2}\\right)$$\n\n$$= 1 + \\frac{2x}{(x^2 + 6)^2}$$\n\nUsing negative indices, we can write the final answer as:\n\n$$1 + 2x(x^2 + 6)^{-2}$$\n\nThis gives you the final accuracy mark **A1**."}]},{"id":"a0baa34c-5846-47cd-9b63-35d326381dfd:single_fraction_quotient","label":"(a) Combining and Using Quotient Rule","steps":[{"type":"text","order":0,"title":"Combine into one fraction","content":"To begin with the **Combining and Using Quotient Rule** method, we first express our function as a single algebraic fraction. We can do this by finding a common denominator:\n\n$$y = x - \\frac{1}{x^2 + 6}$$\n\n$$y = \\frac{x(x^2 + 6)}{x^2 + 6} - \\frac{1}{x^2 + 6}$$\n\n$$y = \\frac{x^3 + 6x - 1}{x^2 + 6}$$. "},{"type":"text","order":1,"title":"Identify u and v","content":"Now we can apply the quotient rule, which is found in your formulas for differentiation. If $y = \\frac{u}{v}$, then:\n\n$$\\frac{dy}{dx} = \\frac{u'v - uv'}{v^2}$$\n\nLet's identify our numerator $u$ and denominator $v$ and differentiate them both with respect to $x$:\n\n- $u = x^3 + 6x - 1 \\implies u' = 3x^2 + 6$\n- $v = x^2 + 6 \\implies v' = 2x$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{\\htmlData{anchor=u-prime-val}{\\color{@sky}{u'}} \\htmlData{anchor=v-val}{\\color{@amber}{v}} - \\htmlData{anchor=u-val}{\\color{@purple}{u}} \\htmlData{anchor=v-prime-val}{\\color{@teal}{v'}}}{\\left(\\htmlData{anchor=v-denom}{\\color{@amber}{v}}\\right)^2}$$"},{"latex":"$$$\\frac{\\htmlData{anchor=u-prime-sub}{\\color{@sky}{(3x^2 + 6)}}\\htmlData{anchor=v-sub}{\\color{@amber}{(x^2 + 6)}} - \\htmlData{anchor=u-sub}{\\color{@purple}{(x^3 + 6x - 1)}}\\htmlData{anchor=v-prime-sub}{\\color{@teal}{(2x)}}}{\\left(\\htmlData{anchor=v-denom-sub}{\\color{@amber}{x^2 + 6}}\\right)^2}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-prime-sub"},{"color":"amber","style":"box","anchor":"v-sub"},{"color":"purple","style":"box","anchor":"u-sub"},{"color":"teal","style":"box","anchor":"v-prime-sub"}],"connections":[{"to":"u-prime-sub","from":"u-prime-val","color":"sky","label":"u'","style":"solid"},{"to":"v-sub","from":"v-val","color":"amber","label":"v","style":"solid"},{"to":"u-sub","from":"u-val","color":"purple","label":"u","style":"solid"},{"to":"v-prime-sub","from":"v-prime-val","color":"teal","label":"v'","style":"solid"}]},"order":2,"title":"Substitute into quotient rule","description":"Substitute our expressions for $u, u', v,$ and $v'$ directly into the quotient rule formula."},{"type":"text","order":3,"title":"Simplify the numerator","content":"Now we expand and simplify the terms in the numerator:\n\n1. Expand the first term:\n\n$$(3x^2 + 6)(x^2 + 6) = 3x^4 + 18x^2 + 6x^2 + 36 = 3x^4 + 24x^2 + 36$$\n\n2. Expand the second term:\n\n$$(x^3 + 6x - 1)(2x) = 2x^4 + 12x^2 - 2x$$\n\n3. Subtract them:\n\n$$\\text{Numerator} = (3x^4 + 24x^2 + 36) - (2x^4 + 12x^2 - 2x)$$\n\n$$= 3x^4 - 2x^4 + 24x^2 - 12x^2 + 2x + 36$$\n\n$$= x^4 + 12x^2 + 2x + 36$$\n\nPutting it all back together over the denominator, we get:\n\n$$\\frac{dy}{dx} = \\frac{x^4 + 12x^2 + 2x + 36}{(x^2 + 6)^2}$$"},{"type":"text","order":4,"title":"Relate to alternative form","content":"The answer we obtained is fully correct! We can also show it is equivalent to the expression in the markscheme, $1 + 2x(x^2 + 6)^{-2}$, by splitting our fraction:\n\n$$\\frac{x^4 + 12x^2 + 2x + 36}{(x^2 + 6)^2} = \\frac{(x^4 + 12x^2 + 36) + 2x}{(x^2 + 6)^2}$$\n\n$$= \\frac{(x^2 + 6)^2 + 2x}{(x^2 + 6)^2}$$\n\n$$= \\frac{(x^2 + 6)^2}{(x^2 + 6)^2} + \\frac{2x}{(x^2 + 6)^2}$$\n\n$$= 1 + \\frac{2x}{(x^2 + 6)^2}$$\n\n$$= 1 + 2x(x^2 + 6)^{-2}$$\n\nBoth forms are completely valid, but combining first into a single fraction is a very robust way to avoid mistakes with negative power indices!"}]},{"id":"e9b524cc-b698-4e5b-ab9d-37ac24370b72:direct_chain_rule","label":"(b) Direct Chain Rule","steps":[{"type":"text","order":0,"title":"Analyze the function","content":"We need to find the derivative of the expression with respect to $x$:\n\n$$\\frac{d}{dx}\\left(2 + e^{0.5x^2}\\right)$$\n\nThis expression is composed of two parts: a constant term, $2$, and an exponential term, $e^{0.5x^2}$. We can differentiate each term individually."},{"type":"text","order":1,"title":"Differentiate the constant","content":"The derivative of any constant term with respect to $x$ is zero:\n\n$$\\frac{d}{dx}(2) = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"For the exponential term, we first identify the exponent:"},{"latex":"$$e^{\\htmlData{anchor=exponent}{\\color{@sky}{0.5x^2}}}$"},{"text":"We then differentiate this exponent with respect to $x$:"},{"latex":"$$$\\frac{d}{dx}(\\htmlData{anchor=exponent-sub}{\\color{@sky}{0.5x^2}}) = \\htmlData{anchor=deriv-exponent}{\\color{@amber}{x}}$$"},{"text":"Finally, we multiply the derivative of the exponent by the original exponential term:"},{"latex":"$$$\\frac{d}{dx}\\left(e^{\\color{@sky}{0.5x^2}}\\right) = \\htmlData{anchor=result-deriv}{\\color{@amber}{x}} e^{\\color{@sky}{0.5x^2}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"amber","style":"text-color","anchor":"deriv-exponent"}],"connections":[{"to":"exponent-sub","from":"exponent","color":"sky","label":null,"style":"solid"},{"to":"result-deriv","from":"deriv-exponent","color":"amber","label":"derivative of exponent","style":"solid"}]},"order":2,"title":"Apply direct chain rule","description":"Using the chain rule, $\\frac{d}{dx}(e^{u}) = u' e^{u}$, we find the derivative of the inner exponent function and multiply it by the original exponential expression."},{"type":"text","order":3,"title":"State the final answer","content":"Combining the derivatives of both terms gives the final result:\n\n$$\\frac{d}{dx}\\left(2 + e^{0.5x^2}\\right) = 0 + x e^{0.5x^2} = x e^{0.5x^2}$$\n\nThis earns the **M1** mark for correctly differentiating the exponential part and the **A1** mark for the correct final simplified expression."}]},{"id":"e9b524cc-b698-4e5b-ab9d-37ac24370b72:formal_substitution","label":"(b) Formal u-Substitution","steps":[{"type":"text","order":0,"title":"Define the target function","content":"Let our function be defined as:\n\n$$y = 2 + e^{0.5x^2}$$\n\nWe want to find the derivative $\\frac{dy}{dx}$. \n\nSince this function contains a constant and a composite function, we can differentiate each term individually. The constant term $2$ will differentiate to $0$. To find the derivative of the exponential term $e^{0.5x^2}$, we will use the **Formal $u$-Substitution** method."},{"type":"hyper_latex","block":{"nodes":[{"text":"We define a new variable for the exponent:"},{"latex":"$$u = \\htmlData{anchor=exponent}{\\color{@sky}{0.5x^2}}$"},{"text":"This allows us to write the function $y$ in terms of $u$:"},{"latex":"$$y = 2 + e^{\\htmlData{anchor=u-var}{\\color{@sky}{u}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"sky","style":"text-color","anchor":"u-var"}],"connections":[{"to":"u-var","from":"exponent","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We identify the inner function as the exponent of the natural exponential function and substitute it with a new variable $u$."},{"type":"text","order":2,"title":"Calculate the individual derivatives","content":"Next, we need to find the derivative of $u$ with respect to $x$, and the derivative of $y$ with respect to $u$.\n\n1. Differentiate $u = 0.5x^2$ with respect to $x$ using the power rule:\n $$\\frac{du}{dx} = 0.5 \\cdot 2x = x$$\n\n2. Differentiate $y = 2 + e^u$ with respect to $u$:\n $$\\frac{dy}{du} = e^u$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Now we apply the formal chain rule formula:"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dy-du}{\\color{@teal}{\\frac{dy}{du}}} \\cdot \\htmlData{anchor=du-dx}{\\color{@purple}{\\frac{du}{dx}}}$"},{"text":"Substituting our expressions into this formula:"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=dy-du-val}{\\color{@teal}{e^u}} \\cdot \\htmlData{anchor=du-dx-val}{\\color{@purple}{x}}$"}],"highlights":[{"color":"teal","style":"text-color","anchor":"dy-du"},{"color":"purple","style":"text-color","anchor":"du-dx"},{"color":"teal","style":"text-color","anchor":"dy-du-val"},{"color":"purple","style":"text-color","anchor":"du-dx-val"}],"connections":[{"to":"dy-du","from":"dy-du-val","color":"teal","label":"substitute dy/du","style":"solid"},{"to":"du-dx","from":"du-dx-val","color":"purple","label":"substitute du/dx","style":"solid"}]},"order":3,"title":"Apply the Chain Rule","description":"We substitute our derivatives into the Chain Rule formula to find $\\frac{dy}{dx}$ in terms of $x$ and $u$."},{"type":"text","order":4,"title":"Substitute back for x","content":"Finally, we substitute our original expression $u = 0.5x^2$ back into the derivative:\n\n$$\\frac{dy}{dx} = x e^{0.5x^2}$$\n\nThis earns both marks: \n- **M1** for correctly differentiating the exponential term using the chain rule.\n- **A1** for the final simplified answer."}]}],"generatedAt":"2026-06-09T09:15:49.644Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"0f10c9b7-4000-4a5f-a375-8b859f75b900","revisionId":"fcb7ca9a-118a-4f98-a753-7bef83e72b2b","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"easy","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"bbcf9741-f67b-434c-8f4a-606c38757a14","label":null,"content":"The depth of water in an experimental filtration vessel, $D$ centimetres, varies according to the radius of the sediment layer $r$ millimetres, such that $D = (2r^3 + 9r^2)^4$. Determine the rate of change of the depth with respect to the radius, $\\frac{dD}{dr}$.","markscheme":"- Identifies the need for the chain rule, for example by letting $u = 2r^3 + 9r^2$ M1\n- Differentiates the inner function to obtain $\\frac{du}{dr} = 6r^2 + 18r$ A1\n- Applies the chain rule formula $\\frac{dD}{dr} = \\frac{dD}{du} \\times \\frac{du}{dr}$ and substitutes $u$ back to find $\\frac{dD}{dr} = 4(2r^3 + 9r^2)^3 \\times (6r^2 + 18r)$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"chain_rule_substitution","label":"Chain Rule with Substitution","steps":[{"type":"text","order":0,"title":"Define substitution variable","content":"To differentiate $D = (2r^3 + 9r^2)^4$, we can use the **chain rule with substitution**.\n\nLet's define an intermediate variable $u$ to represent the inner function:\n$$u = 2r^3 + 9r^2$$\n\nSubstituting this back into the original expression gives $D$ as a function of $u$:\n$$D = u^4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"u = \\color{@sky}{2r^3 + 9r^2} \\implies \\htmlData{anchor=dudr}{\\color{@sky}{\\frac{du}{dr}}} = \\color{@sky}{6r^2 + 18r}"},{"latex":"D = \\color{@amber}{u^4} \\implies \\htmlData{anchor=dddu}{\\color{@amber}{\\frac{dD}{du}}} = \\color{@amber}{4u^3}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dudr"},{"color":"amber","style":"text-color","anchor":"dddu"}],"connections":[]},"order":1,"title":"Differentiate both functions","description":"Now we differentiate each function with respect to its variable using the power rule."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\frac{dD}{dr} = \\htmlData{anchor=dddu-sub}{\\color{@amber}{\\frac{dD}{du}}} \\times \\htmlData{anchor=dudr-sub}{\\color{@sky}{\\frac{du}{dr}}}"},{"latex":"\\frac{dD}{dr} = \\htmlData{anchor=combined}{\\color{@amber}{4u^3} \\times \\color{@sky}{(6r^2 + 18r)}}"}],"highlights":[{"color":"amber","style":"text-color","anchor":"dddu-sub"},{"color":"sky","style":"text-color","anchor":"dudr-sub"}],"connections":[{"to":"combined","from":"dddu-sub","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Apply the chain rule","description":"The chain rule states that to find the rate of change of $D$ with respect to $r$, we multiply these two derivatives together."},{"type":"text","order":3,"title":"Substitute back for u","content":"Finally, we replace $u$ with our original substitution $u = 2r^3 + 9r^2$ to write the derivative purely in terms of $r$:\n\n$$\\frac{dD}{dr} = 4(2r^3 + 9r^2)^3 (6r^2 + 18r)$$\n\nThis is our final derivative expression."}]},{"id":"direct_chain_rule","label":"Direct Chain Rule","steps":[{"type":"text","order":0,"title":"Identify composite function","content":"We need to find the rate of change of the depth with respect to the radius, which is represented by the derivative $\\frac{dD}{dr}$.\n\nThe given equation is:\n$$D = (2r^3 + 9r^2)^4$$\n\nThis is a composite function of the form $D = [f(r)]^4$, where the inner function is $2r^3 + 9r^2$ and the outer function is raising to the power of $4$. We can apply the **Direct Chain Rule** to differentiate it directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"D = (\\htmlData{anchor=inner}{\\color{@sky}{2r^3 + 9r^2}})^{\\htmlData{anchor=power}{\\color{@amber}{4}}}"},{"latex":"\\frac{dD}{dr} = \\htmlData{anchor=power-out}{\\color{@amber}{4}}(\\htmlData{anchor=inner-out}{\\color{@sky}{2r^3 + 9r^2}})^{\\color{@amber}{3}} \\times \\frac{d}{dr}(\\color{@sky}{2r^3 + 9r^2})"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner"},{"color":"amber","style":"text-color","anchor":"power"}],"connections":[{"to":"power-out","from":"power","color":"amber","label":"bring down power","style":"solid"},{"to":"inner-out","from":"inner","color":"sky","label":"keep inner same","style":"solid"}]},"order":1,"title":"Differentiate outer function","description":"According to the general power rule, we differentiate the outer power function first while keeping the inner term exactly the same, and then multiply by the derivative of the inner function."},{"type":"text","order":2,"title":"Differentiate inner function","content":"Now, we find the derivative of the inner function, $2r^3 + 9r^2$, using the basic power rule:\n\n$$\\frac{d}{dr}(2r^3 + 9r^2) = (3 \\times 2)r^2 + (2 \\times 9)r = 6r^2 + 18r$$\n\nThis step earns you the second mark in the exam for correctly finding the derivative of the inner bracket."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\frac{dD}{dr} = 4(2r^3 + 9r^2)^3 \\times \\htmlData{anchor=inner-deriv}{\\color{@green}{(6r^2 + 18r)}}"},{"latex":"\\frac{dD}{dr} = \\htmlData{anchor=final-ans}{\\color{@green}{4(6r^2 + 18r)(2r^3 + 9r^2)^3}}"}],"highlights":[{"color":"green","style":"text-color","anchor":"inner-deriv"}],"connections":[{"to":"final-ans","from":"inner-deriv","color":"green","label":"simplified form","style":"solid"}]},"order":3,"title":"Multiply results together","description":"Combine the results by substituting the derivative of the inner function back into our chain rule expression. This completes the differentiation and earns you the final mark."}]}],"generatedAt":"2026-06-09T03:11:24.831Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"bbcf9741-f67b-434c-8f4a-606c38757a14:chain_rule_substitution","label":"Chain Rule with Substitution","steps":[{"type":"text","order":0,"title":"Define substitution variable","content":"To differentiate $D = (2r^3 + 9r^2)^4$, we can use the **chain rule with substitution**.\n\nLet's define an intermediate variable $u$ to represent the inner function:\n$$u = 2r^3 + 9r^2$$\n\nSubstituting this back into the original expression gives $D$ as a function of $u$:\n$$D = u^4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"u = \\color{@sky}{2r^3 + 9r^2} \\implies \\htmlData{anchor=dudr}{\\color{@sky}{\\frac{du}{dr}}} = \\color{@sky}{6r^2 + 18r}"},{"latex":"D = \\color{@amber}{u^4} \\implies \\htmlData{anchor=dddu}{\\color{@amber}{\\frac{dD}{du}}} = \\color{@amber}{4u^3}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dudr"},{"color":"amber","style":"text-color","anchor":"dddu"}],"connections":[]},"order":1,"title":"Differentiate both functions","description":"Now we differentiate each function with respect to its variable using the power rule."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\frac{dD}{dr} = \\htmlData{anchor=dddu-sub}{\\color{@amber}{\\frac{dD}{du}}} \\times \\htmlData{anchor=dudr-sub}{\\color{@sky}{\\frac{du}{dr}}}"},{"latex":"\\frac{dD}{dr} = \\htmlData{anchor=combined}{\\color{@amber}{4u^3} \\times \\color{@sky}{(6r^2 + 18r)}}"}],"highlights":[{"color":"amber","style":"text-color","anchor":"dddu-sub"},{"color":"sky","style":"text-color","anchor":"dudr-sub"}],"connections":[{"to":"combined","from":"dddu-sub","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Apply the chain rule","description":"The chain rule states that to find the rate of change of $D$ with respect to $r$, we multiply these two derivatives together."},{"type":"text","order":3,"title":"Substitute back for u","content":"Finally, we replace $u$ with our original substitution $u = 2r^3 + 9r^2$ to write the derivative purely in terms of $r$:\n\n$$\\frac{dD}{dr} = 4(2r^3 + 9r^2)^3 (6r^2 + 18r)$$\n\nThis is our final derivative expression."}]},{"id":"bbcf9741-f67b-434c-8f4a-606c38757a14:direct_chain_rule","label":"Direct Chain Rule","steps":[{"type":"text","order":0,"title":"Identify composite function","content":"We need to find the rate of change of the depth with respect to the radius, which is represented by the derivative $\\frac{dD}{dr}$.\n\nThe given equation is:\n$$D = (2r^3 + 9r^2)^4$$\n\nThis is a composite function of the form $D = [f(r)]^4$, where the inner function is $2r^3 + 9r^2$ and the outer function is raising to the power of $4$. We can apply the **Direct Chain Rule** to differentiate it directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"D = (\\htmlData{anchor=inner}{\\color{@sky}{2r^3 + 9r^2}})^{\\htmlData{anchor=power}{\\color{@amber}{4}}}"},{"latex":"\\frac{dD}{dr} = \\htmlData{anchor=power-out}{\\color{@amber}{4}}(\\htmlData{anchor=inner-out}{\\color{@sky}{2r^3 + 9r^2}})^{\\color{@amber}{3}} \\times \\frac{d}{dr}(\\color{@sky}{2r^3 + 9r^2})"}],"highlights":[{"color":"sky","style":"text-color","anchor":"inner"},{"color":"amber","style":"text-color","anchor":"power"}],"connections":[{"to":"power-out","from":"power","color":"amber","label":"bring down power","style":"solid"},{"to":"inner-out","from":"inner","color":"sky","label":"keep inner same","style":"solid"}]},"order":1,"title":"Differentiate outer function","description":"According to the general power rule, we differentiate the outer power function first while keeping the inner term exactly the same, and then multiply by the derivative of the inner function."},{"type":"text","order":2,"title":"Differentiate inner function","content":"Now, we find the derivative of the inner function, $2r^3 + 9r^2$, using the basic power rule:\n\n$$\\frac{d}{dr}(2r^3 + 9r^2) = (3 \\times 2)r^2 + (2 \\times 9)r = 6r^2 + 18r$$\n\nThis step earns you the second mark in the exam for correctly finding the derivative of the inner bracket."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\frac{dD}{dr} = 4(2r^3 + 9r^2)^3 \\times \\htmlData{anchor=inner-deriv}{\\color{@green}{(6r^2 + 18r)}}"},{"latex":"\\frac{dD}{dr} = \\htmlData{anchor=final-ans}{\\color{@green}{4(6r^2 + 18r)(2r^3 + 9r^2)^3}}"}],"highlights":[{"color":"green","style":"text-color","anchor":"inner-deriv"}],"connections":[{"to":"final-ans","from":"inner-deriv","color":"green","label":"simplified form","style":"solid"}]},"order":3,"title":"Multiply results together","description":"Combine the results by substituting the derivative of the inner function back into our chain rule expression. This completes the differentiation and earns you the final 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Chain rule (A-level only)","topicId":"eab581f5-8603-43fa-a9d3-b6c87c0d8796","topicSidebar":"$L3b","topicSlug":"chain-rule","topicTitle":"1.9.14 Chain rule (A-level only)"}]}],"$L3c","$L3d"]}]

1.9.14 Chain rule (A-level only)

Question 7

1.9.14 Chain rule (A-level only) Questions

1.9.14 Chain rule (A-level only)

Question 7

1.9.14 Chain rule (A-level only) Questions