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bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-9-4-the-product-rule/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":21,"initialQuestionWindow":[{"id":"628b555b-bb68-4df3-9cd9-fee55098c4c7","revisionId":"70902d46-0f1b-46a8-ac1e-0fdd4441b3e8","questionNumber":null,"questionType":"LA","specification":"A scientist models the activity, \$A\$ MBq, of a radioactive isotope over time, \$t\$ days, using the equation\n\n\\[A = (15t + 2)e^{-0.4t}, \\quad t \\ge 0\\]","difficultyBand":"easy","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"79ec8cab-fee2-4c5f-8e46-b1c5c08fa63b","label":null,"content":"Find, using calculus, the exact value of \$t\$ for which the activity is stationary.","markscheme":"- Recognises the need for the product rule and identifies $u = 15t + 2$ and $v = e^{-0.4t}$ with derivatives $u' = 15$ and $v' = -0.4e^{-0.4t}$ M1\n- Correctly applies the product rule to obtain $\\frac{dA}{dt} = (15t + 2)(-0.4e^{-0.4t}) + 15e^{-0.4t}$ M1\n- Sets $\\frac{dA}{dt} = 0$ and factors out the exponential term to reach $15 - 0.4(15t + 2) = 0$ or $14.2 - 6t = 0$ M1\n- Solves the linear equation to find the exact value $t = \\frac{71}{30}$ or equivalent exact fraction A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"product_rule","label":"Using the Product Rule","steps":[{"type":"text","order":0,"title":"Identify the functions","content":"To find the stationary point of the activity $A = (15t + 2)e^{-0.4t}$, we need to find the derivative $\\frac{dA}{dt}$ and set it equal to $0$.\n\nSince $A$ is the product of two functions of $t$, we will use the product rule:\n\n$$\\frac{d}{dt}(uv) = u\\frac{dv}{dt} + v\\frac{du}{dt}$$\n\nLet's define our two functions $u$ and $v$:\n\n$$\\begin{aligned} u &= 15t + 2 \\\\ v &= e^{-0.4t} \\end{aligned}$$"},{"type":"text","order":1,"title":"Differentiate each part","content":"We differentiate each function with respect to $t$. For $v$, we apply the chain rule:\n\nFor $u = 15t + 2$:\n$$\\frac{du}{dt} = 15$$\n\nFor $v = e^{-0.4t}$:\n$$\\frac{dv}{dt} = -0.4e^{-0.4t}$$"},{"type":"text","order":2,"title":"Apply the product rule","content":"Now we substitute our functions and their derivatives into the product rule formula:\n\n$$\\frac{dA}{dt} = u\\frac{dv}{dt} + v\\frac{du}{dt}$$\n\n$$\\frac{dA}{dt} = (15t + 2)\\left(-0.4e^{-0.4t}\\right) + 15e^{-0.4t}$$\n\nApplying this correctly earns you the first two marks (**M1 M1**)."},{"type":"text","order":3,"title":"Set the derivative to zero","content":"For a stationary value, we set the rate of change $\\frac{dA}{dt} = 0$:\n\n$$(15t + 2)\\left(-0.4e^{-0.4t}\\right) + 15e^{-0.4t} = 0$$\n\nTo solve this, we can factor out the common exponential term $e^{-0.4t}$:\n\n$$e^{-0.4t} \\left( -0.4(15t + 2) + 15 \\right) = 0$$\n\nSetting up this equation earns the third mark (**M1**)."},{"type":"text","order":4,"title":"Solve the linear equation","content":"Since $e^{-0.4t} \\neq 0$ for all real values of $t$, we can divide both sides of the equation by $e^{-0.4t}$. This leaves us with a linear equation to solve:\n\n$$-0.4(15t + 2) + 15 = 0$$\n\nLet's expand and simplify:\n\n$$\\begin{aligned} -6t - 0.8 + 15 &= 0 \\\\ 14.2 - 6t &= 0 \\\\ 6t &= 14.2 \\end{aligned}$$\n\nConvert the decimal to a fraction to find the exact value of $t$:\n\n$$t = \\frac{14.2}{6} = \\frac{142}{60} = \\frac{71}{30}$$\n\nThis gives the final accuracy mark (**A1**).\n\nThus, the exact value of $t$ for which the activity is stationary is:\n\n$$t = \\frac{71}{30} \\text{ days}$$\n\n*(Note: This is approximately $2.37$ days, but the question specifically asks for the exact value.)*"}]},{"id":"quotient_rule","label":"Using the Quotient Rule","steps":[{"type":"text","order":0,"title":"Rewrite as a fraction","content":"To use the quotient rule, we first rewrite the negative exponential term as a division. \n\nRecall that $e^{-x} = \\frac{1}{e^x}$. We can apply this to rewrite the activity equation:\n\n$$A = \\frac{15t + 2}{e^{0.4t}}$$\n\nNow, this function is in the form of a quotient, $A = \\frac{u}{v}$, which allows us to apply the quotient rule directly."},{"type":"text","order":1,"title":"Identify terms and derivatives","content":"We define our numerator $u$ and denominator $v$:\n\n$$u = 15t + 2 \\quad \\text{and} \\quad v = e^{0.4t}$$\n\nNow, we find their individual derivatives with respect to $t$:\n\n$$\\frac{du}{dt} = 15 \\quad \\text{and} \\quad \\frac{dv}{dt} = 0.4e^{0.4t}$$"},{"type":"text","order":2,"title":"Apply the quotient rule","content":"The quotient rule is given by:\n\n$$\\frac{dA}{dt} = \\frac{v\\frac{du}{dt} - u\\frac{dv}{dt}}{v^2}$$\n\nSubstituting our expressions for $u, v, \\frac{du}{dt},$ and $\\frac{dv}{dt}$ into the formula:\n\n$$\\frac{dA}{dt} = \\frac{(e^{0.4t})(15) - (15t + 2)(0.4e^{0.4t})}{(e^{0.4t})^2}$$\n\nThis expression would earn you the first two marks under a standard product/quotient rule differentiation scheme."},{"type":"text","order":3,"title":"Simplify the derivative","content":"To simplify the numerator, we can factor out $e^{0.4t}$:\n\n$$\\frac{dA}{dt} = \\frac{e^{0.4t} \\left( 15 - 0.4(15t + 2) \\right)}{(e^{0.4t})^2}$$\n\nWe can cancel one factor of $e^{0.4t}$ from the numerator and denominator:\n\n$$\\frac{dA}{dt} = \\frac{15 - 0.4(15t + 2)}{e^{0.4t}}$$\n\nNow, expand and simplify the terms in the numerator:\n\n$$15 - 0.4(15t + 2) = 15 - 6t - 0.8 = 14.2 - 6t$$\n\nThus, the simplified derivative is:\n\n$$\\frac{dA}{dt} = \\frac{14.2 - 6t}{e^{0.4t}}$$"},{"type":"text","order":4,"title":"Solve for stationary points","content":"A stationary point occurs when the derivative is equal to zero:\n\n$$\\frac{dA}{dt} = 0 \\quad \\Rightarrow \\quad \\frac{14.2 - 6t}{e^{0.4t}} = 0$$\n\nSince an exponential term $e^{0.4t}$ is strictly positive for all real values of $t$, a fraction can only be zero if its numerator is zero:\n\n$$14.2 - 6t = 0$$\n\nSolving for $t$:\n\n$$6t = 14.2$$\n\n$$t = \\frac{14.2}{6} = \\frac{142}{60} = \\frac{71}{30}$$\n\nSince the question asks for the **exact value**, we must leave our final answer as a simplified fraction:\n\n$$t = \\frac{71}{30} \\text{ days}$$"}]},{"id":"logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Introduce logarithmic differentiation","content":"To find the stationary value of $t$, we need to find when the rate of change of activity is zero, $\\frac{dA}{dt} = 0$.\n\nInstead of the standard product rule, we can simplify the differentiation of products and exponentials by using **logarithmic differentiation**.\n\nFirst, we take the natural logarithm, $\\ln$, of both sides of the activity equation:\n\n$$\\ln(A) = \\ln\\left((15t + 2)e^{-0.4t}\\right)$$"},{"type":"text","order":1,"title":"Apply logarithm laws","content":"Using the product rule for logarithms, $\\ln(xy) = \\ln(x) + \\ln(y)$, we can expand the right-hand side:\n\n$$\\ln(A) = \\ln(15t + 2) + \\ln\\left(e^{-0.4t}\\right)$$\n\nSince $\\ln\\left(e^x\\right) = x$, we can simplify the second term to obtain:\n\n$$\\ln(A) = \\ln(15t + 2) - 0.4t$$"},{"type":"text","order":2,"title":"Differentiate implicitly","content":"Now, we differentiate both sides with respect to $t$. \n\nRecall that by the chain rule, the derivative of $\\ln(A)$ with respect to $t$ is:\n\n$$\\frac{d}{dt}\\left[\\ln(A)\\right] = \\frac{1}{A}\\frac{dA}{dt}$$\n\nDifferentiating the right-hand side using the chain rule for $\\ln(15t + 2)$:\n\n$$\\frac{1}{A}\\frac{dA}{dt} = \\frac{15}{15t + 2} - 0.4$$"},{"type":"text","order":3,"title":"Set derivative to zero","content":"For a stationary point, we set the rate of change of activity to zero: \n\n$$\\frac{dA}{dt} = 0$$\n\nSubstituting this into our differentiated equation:\n\n$$\\frac{1}{A}(0) = \\frac{15}{15t + 2} - 0.4$$\n\n$$0 = \\frac{15}{15t + 2} - 0.4$$"},{"type":"text","order":4,"title":"Solve for t","content":"Now, we solve this linear equation for $t$ to find the exact time of the stationary point:\n\n$$\\begin{aligned} 0.4 &= \\frac{15}{15t + 2} \\\\ 0.4(15t + 2) &= 15 \\\\ 6t + 0.8 &= 15 \\\\ 6t &= 14.2 \\\\ t &= \\frac{14.2}{6} \\end{aligned}$$\n\nTo write this as an exact fraction, we multiply the numerator and the denominator by $10$:\n\n$$t = \\frac{142}{60} = \\frac{71}{30}$$"}]}],"generatedAt":"2026-06-09T06:02:19.278Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"79ec8cab-fee2-4c5f-8e46-b1c5c08fa63b:product_rule","label":"Using the Product Rule","steps":[{"type":"text","order":0,"title":"Identify the functions","content":"To find the stationary point of the activity $A = (15t + 2)e^{-0.4t}$, we need to find the derivative $\\frac{dA}{dt}$ and set it equal to $0$.\n\nSince $A$ is the product of two functions of $t$, we will use the product rule:\n\n$$\\frac{d}{dt}(uv) = u\\frac{dv}{dt} + v\\frac{du}{dt}$$\n\nLet's define our two functions $u$ and $v$:\n\n$$\\begin{aligned} u &= 15t + 2 \\\\ v &= e^{-0.4t} \\end{aligned}$$"},{"type":"text","order":1,"title":"Differentiate each part","content":"We differentiate each function with respect to $t$. For $v$, we apply the chain rule:\n\nFor $u = 15t + 2$:\n$$\\frac{du}{dt} = 15$$\n\nFor $v = e^{-0.4t}$:\n$$\\frac{dv}{dt} = -0.4e^{-0.4t}$$"},{"type":"text","order":2,"title":"Apply the product rule","content":"Now we substitute our functions and their derivatives into the product rule formula:\n\n$$\\frac{dA}{dt} = u\\frac{dv}{dt} + v\\frac{du}{dt}$$\n\n$$\\frac{dA}{dt} = (15t + 2)\\left(-0.4e^{-0.4t}\\right) + 15e^{-0.4t}$$\n\nApplying this correctly earns you the first two marks (**M1 M1**)."},{"type":"text","order":3,"title":"Set the derivative to zero","content":"For a stationary value, we set the rate of change $\\frac{dA}{dt} = 0$:\n\n$$(15t + 2)\\left(-0.4e^{-0.4t}\\right) + 15e^{-0.4t} = 0$$\n\nTo solve this, we can factor out the common exponential term $e^{-0.4t}$:\n\n$$e^{-0.4t} \\left( -0.4(15t + 2) + 15 \\right) = 0$$\n\nSetting up this equation earns the third mark (**M1**)."},{"type":"text","order":4,"title":"Solve the linear equation","content":"Since $e^{-0.4t} \\neq 0$ for all real values of $t$, we can divide both sides of the equation by $e^{-0.4t}$. This leaves us with a linear equation to solve:\n\n$$-0.4(15t + 2) + 15 = 0$$\n\nLet's expand and simplify:\n\n$$\\begin{aligned} -6t - 0.8 + 15 &= 0 \\\\ 14.2 - 6t &= 0 \\\\ 6t &= 14.2 \\end{aligned}$$\n\nConvert the decimal to a fraction to find the exact value of $t$:\n\n$$t = \\frac{14.2}{6} = \\frac{142}{60} = \\frac{71}{30}$$\n\nThis gives the final accuracy mark (**A1**).\n\nThus, the exact value of $t$ for which the activity is stationary is:\n\n$$t = \\frac{71}{30} \\text{ days}$$\n\n*(Note: This is approximately $2.37$ days, but the question specifically asks for the exact value.)*"}]},{"id":"79ec8cab-fee2-4c5f-8e46-b1c5c08fa63b:quotient_rule","label":"Using the Quotient Rule","steps":[{"type":"text","order":0,"title":"Rewrite as a fraction","content":"To use the quotient rule, we first rewrite the negative exponential term as a division. \n\nRecall that $e^{-x} = \\frac{1}{e^x}$. We can apply this to rewrite the activity equation:\n\n$$A = \\frac{15t + 2}{e^{0.4t}}$$\n\nNow, this function is in the form of a quotient, $A = \\frac{u}{v}$, which allows us to apply the quotient rule directly."},{"type":"text","order":1,"title":"Identify terms and derivatives","content":"We define our numerator $u$ and denominator $v$:\n\n$$u = 15t + 2 \\quad \\text{and} \\quad v = e^{0.4t}$$\n\nNow, we find their individual derivatives with respect to $t$:\n\n$$\\frac{du}{dt} = 15 \\quad \\text{and} \\quad \\frac{dv}{dt} = 0.4e^{0.4t}$$"},{"type":"text","order":2,"title":"Apply the quotient rule","content":"The quotient rule is given by:\n\n$$\\frac{dA}{dt} = \\frac{v\\frac{du}{dt} - u\\frac{dv}{dt}}{v^2}$$\n\nSubstituting our expressions for $u, v, \\frac{du}{dt},$ and $\\frac{dv}{dt}$ into the formula:\n\n$$\\frac{dA}{dt} = \\frac{(e^{0.4t})(15) - (15t + 2)(0.4e^{0.4t})}{(e^{0.4t})^2}$$\n\nThis expression would earn you the first two marks under a standard product/quotient rule differentiation scheme."},{"type":"text","order":3,"title":"Simplify the derivative","content":"To simplify the numerator, we can factor out $e^{0.4t}$:\n\n$$\\frac{dA}{dt} = \\frac{e^{0.4t} \\left( 15 - 0.4(15t + 2) \\right)}{(e^{0.4t})^2}$$\n\nWe can cancel one factor of $e^{0.4t}$ from the numerator and denominator:\n\n$$\\frac{dA}{dt} = \\frac{15 - 0.4(15t + 2)}{e^{0.4t}}$$\n\nNow, expand and simplify the terms in the numerator:\n\n$$15 - 0.4(15t + 2) = 15 - 6t - 0.8 = 14.2 - 6t$$\n\nThus, the simplified derivative is:\n\n$$\\frac{dA}{dt} = \\frac{14.2 - 6t}{e^{0.4t}}$$"},{"type":"text","order":4,"title":"Solve for stationary points","content":"A stationary point occurs when the derivative is equal to zero:\n\n$$\\frac{dA}{dt} = 0 \\quad \\Rightarrow \\quad \\frac{14.2 - 6t}{e^{0.4t}} = 0$$\n\nSince an exponential term $e^{0.4t}$ is strictly positive for all real values of $t$, a fraction can only be zero if its numerator is zero:\n\n$$14.2 - 6t = 0$$\n\nSolving for $t$:\n\n$$6t = 14.2$$\n\n$$t = \\frac{14.2}{6} = \\frac{142}{60} = \\frac{71}{30}$$\n\nSince the question asks for the **exact value**, we must leave our final answer as a simplified fraction:\n\n$$t = \\frac{71}{30} \\text{ days}$$"}]},{"id":"79ec8cab-fee2-4c5f-8e46-b1c5c08fa63b:logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Introduce logarithmic differentiation","content":"To find the stationary value of $t$, we need to find when the rate of change of activity is zero, $\\frac{dA}{dt} = 0$.\n\nInstead of the standard product rule, we can simplify the differentiation of products and exponentials by using **logarithmic differentiation**.\n\nFirst, we take the natural logarithm, $\\ln$, of both sides of the activity equation:\n\n$$\\ln(A) = \\ln\\left((15t + 2)e^{-0.4t}\\right)$$"},{"type":"text","order":1,"title":"Apply logarithm laws","content":"Using the product rule for logarithms, $\\ln(xy) = \\ln(x) + \\ln(y)$, we can expand the right-hand side:\n\n$$\\ln(A) = \\ln(15t + 2) + \\ln\\left(e^{-0.4t}\\right)$$\n\nSince $\\ln\\left(e^x\\right) = x$, we can simplify the second term to obtain:\n\n$$\\ln(A) = \\ln(15t + 2) - 0.4t$$"},{"type":"text","order":2,"title":"Differentiate implicitly","content":"Now, we differentiate both sides with respect to $t$. \n\nRecall that by the chain rule, the derivative of $\\ln(A)$ with respect to $t$ is:\n\n$$\\frac{d}{dt}\\left[\\ln(A)\\right] = \\frac{1}{A}\\frac{dA}{dt}$$\n\nDifferentiating the right-hand side using the chain rule for $\\ln(15t + 2)$:\n\n$$\\frac{1}{A}\\frac{dA}{dt} = \\frac{15}{15t + 2} - 0.4$$"},{"type":"text","order":3,"title":"Set derivative to zero","content":"For a stationary point, we set the rate of change of activity to zero: \n\n$$\\frac{dA}{dt} = 0$$\n\nSubstituting this into our differentiated equation:\n\n$$\\frac{1}{A}(0) = \\frac{15}{15t + 2} - 0.4$$\n\n$$0 = \\frac{15}{15t + 2} - 0.4$$"},{"type":"text","order":4,"title":"Solve for t","content":"Now, we solve this linear equation for $t$ to find the exact time of the stationary point:\n\n$$\\begin{aligned} 0.4 &= \\frac{15}{15t + 2} \\\\ 0.4(15t + 2) &= 15 \\\\ 6t + 0.8 &= 15 \\\\ 6t &= 14.2 \\\\ t &= \\frac{14.2}{6} \\end{aligned}$$\n\nTo write this as an exact fraction, we multiply the numerator and the denominator by $10$:\n\n$$t = \\frac{142}{60} = \\frac{71}{30}$$"}]}],"generatedAt":"2026-06-09T06:02:32.611Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"798c117d-a4ff-46de-9f87-692828cce780","revisionId":"e605c3e2-2e85-43cf-8210-01466eba975f","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"easy","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"da57894b-a7b4-41d1-9b18-0b4871a190df","label":null,"content":"Find, using calculus, the \$x\$ coordinate of the stationary point on the curve with equation\n\n\\[y = (4x - 7)e^{5x}\\]","markscheme":"- Differentiates $u = 4x - 7$ to find $\\frac{du}{dx} = 4$ and $v = e^{5x}$ to find $\\frac{dv}{dx} = 5e^{5x}$ M1\n- Applies the product rule to get $\\frac{dy}{dx} = (4x - 7)(5e^{5x}) + (e^{5x})(4)$ A1\n- Sets $\\frac{dy}{dx} = 0$ and factorises to $e^{5x} [5(4x - 7) + 4] = 0$ M1\n- Solves $20x - 31 = 0$ to find $x = \\frac{31}{20}$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"product_rule","label":"Direct Product Rule","steps":[{"type":"text","order":0,"title":"Identify the method","content":"To find the stationary point, we need to find the derivative of the function and set it equal to $0$:\n\n$$\\frac{dy}{dx} = 0$$\n\nSince the equation $y = (4x - 7)e^{5x}$ is a product of two functions, we will apply the **Product Rule**:\n\n$$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$","highlights":[{"text":"stationary point","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u-def}{\\color{@sky}{u = 4x - 7}} \\quad \\text{and} \\quad \\htmlData{anchor=v-def}{\\color{@amber}{v = e^{5x}}}$"},{"text":"Differentiating each term gives:"},{"latex":"$$\\htmlData{anchor=du-def}{\\color{@sky}{\\frac{du}{dx} = 4}} \\quad \\text{and} \\quad \\htmlData{anchor=dv-def}{\\color{@amber}{\\frac{dv}{dx} = 5e^{5x}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"amber","style":"text-color","anchor":"v-def"}],"connections":[{"to":"du-def","from":"u-def","color":"sky","label":"differentiate","style":"solid"},{"to":"dv-def","from":"v-def","color":"amber","label":"differentiate (chain rule)","style":"solid"}]},"order":1,"title":"Define u and v","description":"This earns you the first method mark **M1** on the exam."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=u-sub}{\\color{@sky}{(4x - 7)}}\\htmlData{anchor=dv-sub}{\\color{@amber}{(5e^{5x})}} + \\htmlData{anchor=v-sub}{\\color{@amber}{(e^{5x})}}\\htmlData{anchor=du-sub}{\\color{@sky}{(4)}}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"u-sub"},{"color":"sky","style":"underline","anchor":"du-sub"},{"color":"amber","style":"underline","anchor":"v-sub"},{"color":"amber","style":"underline","anchor":"dv-sub"}],"connections":[]},"order":2,"title":"Apply the product rule","description":"Substituting our terms into the product rule formula gives the expression for $\\frac{dy}{dx}$. This earns you an accuracy mark **A1**."},{"type":"text","order":3,"title":"Factorise the derivative","content":"To solve $\\frac{dy}{dx} = 0$, we set our derivative to $0$ and factorise out the common exponential term, $e^{5x}$:\n\n$$e^{5x} [5(4x - 7) + 4] = 0$$\n\n$$e^{5x} (20x - 35 + 4) = 0$$\n\n$$e^{5x} (20x - 31) = 0$$\n\nThis factorization step is worth a method mark **M1**."},{"type":"text","order":4,"title":"Solve for x","content":"Since the exponential term $e^{5x}$ can never equal $0$ for any real value of $x$, we must have:\n\n$$20x - 31 = 0$$\n\n$$20x = 31$$\n\n$$x = \\frac{31}{20}$$\n\nThus, the $x$-coordinate of the stationary point is $x = \\frac{31}{20}$ (or $1.55$). This final step awards you the accuracy mark **A1**.","calculator":[{"gdc":{"expression":"x = 1.55","instructions":"Graph the original function y = (4x - 7)e^(5x) and use the GDC's 'Minimum' tool to locate the coordinates of the turning point directly."},"ti84":{"expression":"x = 1.55","instructions":"1. Press [Y=].\n2. Enter Y1 = (4X - 7) * e^(5X).\n3. Press [2nd] [TRACE] and select '6: dy/dx' to find the derivative at a point, or graph it to observe the minimum. Alternatively, use '2: zero' on the graph of the derivative Y2 = e^(5X) * (20X - 31) to find the root."},"desmos":{"expression":"(1.55, -2.138)","instructions":"Type the equation y = (4x - 7)e^(5x) into Desmos. Click on the gray point representing the minimum on the curve to see its coordinates."},"description":"We can verify the solution graphically by plotting the derivative and finding its root, or by finding the stationary point of the original function."}]}]},{"id":"logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Introduce absolute logarithmic differentiation","content":"To find the stationary point of $y = (4x - 7)e^{5x}$ using logarithmic differentiation, we must be careful with the sign of $(4x - 7)$. Since the logarithm of a negative number is undefined in the real number system, we take the natural logarithm of the absolute value of both sides:\n\n$$\\ln|y| = \\ln\\left|(4x - 7)e^{5x}\\right|$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\ln|y| = \\ln|4x - 7| + \\htmlData{anchor=log-exp}{\\ln\\left(e^{5x}\\right)}$"},{"latex":"$$\\ln|y| = \\htmlData{anchor=term-linear}{\\ln|4x - 7|} + \\htmlData{anchor=term-exp}{5x}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"log-exp"},{"color":"sky","style":"text-color","anchor":"term-exp"}],"connections":[{"to":"term-exp","from":"log-exp","color":"sky","label":"simplify","style":"solid"}]},"order":1,"title":"Simplify using logarithm laws","description":"Using the properties of logarithms, we split the product into a sum and simplify the exponential term: $\\ln(e^A) = A$."},{"type":"text","order":2,"title":"Differentiate implicitly","content":"Now we differentiate both sides with respect to $x$. Recall that for any differentiable function $u$, the derivative of $\\ln|u|$ is $\\frac{1}{u}\\frac{du}{dx}$.\n\nApplying this chain rule and implicit differentiation, we obtain:\n\n$$\\frac{1}{y}\\frac{dy}{dx} = \\frac{1}{4x - 7} \\cdot 4 + 5$$\n\n$$\\frac{1}{y}\\frac{dy}{dx} = \\frac{4}{4x - 7} + 5$$"},{"type":"text","order":3,"title":"Set derivative to zero","content":"At a stationary point, the first derivative is zero, so we set $\\frac{dy}{dx} = 0$:\n\n$$y \\left( \\frac{4}{4x - 7} + 5 \\right) = 0$$\n\nSince $y = (4x - 7)e^{5x}$, and neither $e^{5x}$ nor $(4x - 7)$ is zero at this point, we must have:\n\n$$\\frac{4}{4x - 7} + 5 = 0$$"},{"type":"text","order":4,"title":"Solve for x","content":"We now solve this linear equation for $x$:\n\n$$\\frac{4}{4x - 7} = -5$$\n\n$$4 = -5(4x - 7)$$\n\n$$4 = -20x + 35$$\n\n$$20x = 31$$\n\n$$x = \\frac{31}{20} = 1.55$$\n\nThis is the $x$-coordinate of the stationary point."}]}],"generatedAt":"2026-06-09T06:31:52.431Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"da57894b-a7b4-41d1-9b18-0b4871a190df:product_rule","label":"Direct Product Rule","steps":[{"type":"text","order":0,"title":"Identify the method","content":"To find the stationary point, we need to find the derivative of the function and set it equal to $0$:\n\n$$\\frac{dy}{dx} = 0$$\n\nSince the equation $y = (4x - 7)e^{5x}$ is a product of two functions, we will apply the **Product Rule**:\n\n$$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$","highlights":[{"text":"stationary point","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u-def}{\\color{@sky}{u = 4x - 7}} \\quad \\text{and} \\quad \\htmlData{anchor=v-def}{\\color{@amber}{v = e^{5x}}}$"},{"text":"Differentiating each term gives:"},{"latex":"$$\\htmlData{anchor=du-def}{\\color{@sky}{\\frac{du}{dx} = 4}} \\quad \\text{and} \\quad \\htmlData{anchor=dv-def}{\\color{@amber}{\\frac{dv}{dx} = 5e^{5x}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"amber","style":"text-color","anchor":"v-def"}],"connections":[{"to":"du-def","from":"u-def","color":"sky","label":"differentiate","style":"solid"},{"to":"dv-def","from":"v-def","color":"amber","label":"differentiate (chain rule)","style":"solid"}]},"order":1,"title":"Define u and v","description":"This earns you the first method mark **M1** on the exam."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=u-sub}{\\color{@sky}{(4x - 7)}}\\htmlData{anchor=dv-sub}{\\color{@amber}{(5e^{5x})}} + \\htmlData{anchor=v-sub}{\\color{@amber}{(e^{5x})}}\\htmlData{anchor=du-sub}{\\color{@sky}{(4)}}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"u-sub"},{"color":"sky","style":"underline","anchor":"du-sub"},{"color":"amber","style":"underline","anchor":"v-sub"},{"color":"amber","style":"underline","anchor":"dv-sub"}],"connections":[]},"order":2,"title":"Apply the product rule","description":"Substituting our terms into the product rule formula gives the expression for $\\frac{dy}{dx}$. This earns you an accuracy mark **A1**."},{"type":"text","order":3,"title":"Factorise the derivative","content":"To solve $\\frac{dy}{dx} = 0$, we set our derivative to $0$ and factorise out the common exponential term, $e^{5x}$:\n\n$$e^{5x} [5(4x - 7) + 4] = 0$$\n\n$$e^{5x} (20x - 35 + 4) = 0$$\n\n$$e^{5x} (20x - 31) = 0$$\n\nThis factorization step is worth a method mark **M1**."},{"type":"text","order":4,"title":"Solve for x","content":"Since the exponential term $e^{5x}$ can never equal $0$ for any real value of $x$, we must have:\n\n$$20x - 31 = 0$$\n\n$$20x = 31$$\n\n$$x = \\frac{31}{20}$$\n\nThus, the $x$-coordinate of the stationary point is $x = \\frac{31}{20}$ (or $1.55$). This final step awards you the accuracy mark **A1**.","calculator":[{"gdc":{"expression":"x = 1.55","instructions":"Graph the original function y = (4x - 7)e^(5x) and use the GDC's 'Minimum' tool to locate the coordinates of the turning point directly."},"ti84":{"expression":"x = 1.55","instructions":"1. Press [Y=].\n2. Enter Y1 = (4X - 7) * e^(5X).\n3. Press [2nd] [TRACE] and select '6: dy/dx' to find the derivative at a point, or graph it to observe the minimum. Alternatively, use '2: zero' on the graph of the derivative Y2 = e^(5X) * (20X - 31) to find the root."},"desmos":{"expression":"(1.55, -2.138)","instructions":"Type the equation y = (4x - 7)e^(5x) into Desmos. Click on the gray point representing the minimum on the curve to see its coordinates."},"description":"We can verify the solution graphically by plotting the derivative and finding its root, or by finding the stationary point of the original function."}]}]},{"id":"da57894b-a7b4-41d1-9b18-0b4871a190df:logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Introduce absolute logarithmic differentiation","content":"To find the stationary point of $y = (4x - 7)e^{5x}$ using logarithmic differentiation, we must be careful with the sign of $(4x - 7)$. Since the logarithm of a negative number is undefined in the real number system, we take the natural logarithm of the absolute value of both sides:\n\n$$\\ln|y| = \\ln\\left|(4x - 7)e^{5x}\\right|$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\ln|y| = \\ln|4x - 7| + \\htmlData{anchor=log-exp}{\\ln\\left(e^{5x}\\right)}$"},{"latex":"$$\\ln|y| = \\htmlData{anchor=term-linear}{\\ln|4x - 7|} + \\htmlData{anchor=term-exp}{5x}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"log-exp"},{"color":"sky","style":"text-color","anchor":"term-exp"}],"connections":[{"to":"term-exp","from":"log-exp","color":"sky","label":"simplify","style":"solid"}]},"order":1,"title":"Simplify using logarithm laws","description":"Using the properties of logarithms, we split the product into a sum and simplify the exponential term: $\\ln(e^A) = A$."},{"type":"text","order":2,"title":"Differentiate implicitly","content":"Now we differentiate both sides with respect to $x$. Recall that for any differentiable function $u$, the derivative of $\\ln|u|$ is $\\frac{1}{u}\\frac{du}{dx}$.\n\nApplying this chain rule and implicit differentiation, we obtain:\n\n$$\\frac{1}{y}\\frac{dy}{dx} = \\frac{1}{4x - 7} \\cdot 4 + 5$$\n\n$$\\frac{1}{y}\\frac{dy}{dx} = \\frac{4}{4x - 7} + 5$$"},{"type":"text","order":3,"title":"Set derivative to zero","content":"At a stationary point, the first derivative is zero, so we set $\\frac{dy}{dx} = 0$:\n\n$$y \\left( \\frac{4}{4x - 7} + 5 \\right) = 0$$\n\nSince $y = (4x - 7)e^{5x}$, and neither $e^{5x}$ nor $(4x - 7)$ is zero at this point, we must have:\n\n$$\\frac{4}{4x - 7} + 5 = 0$$"},{"type":"text","order":4,"title":"Solve for x","content":"We now solve this linear equation for $x$:\n\n$$\\frac{4}{4x - 7} = -5$$\n\n$$4 = -5(4x - 7)$$\n\n$$4 = -20x + 35$$\n\n$$20x = 31$$\n\n$$x = \\frac{31}{20} = 1.55$$\n\nThis is the $x$-coordinate of the stationary point."}]}],"generatedAt":"2026-06-09T06:32:03.042Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"8ebc1394-5cdb-455f-a25d-6766c150b7ba","revisionId":"4119a298-02ac-4c7c-afde-db618e1b7ea1","questionNumber":null,"questionType":"LA","specification":"The displacement, \$s\$ millimetres, of a robotic arm at time \$t\$ seconds is given by the equation\n\n\\[s = (0.35t - 4.2) e^{0.4t}\\]","difficultyBand":"easy","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"f196addb-e240-4981-9709-25479d539507","label":null,"content":"Determine, using calculus, the exact value of \$t\$ for which the displacement has a stationary point.","markscheme":"- Differentiates $0.35t - 4.2$ to obtain $0.35$ and $e^{0.4t}$ to obtain $0.4e^{0.4t}$ B1\n- Applies the product rule to obtain $\\frac{ds}{dt} = 0.35e^{0.4t} + 0.4(0.35t - 4.2)e^{0.4t}$ M1\n- Sets $\\frac{ds}{dt} = 0$ and factorises out $e^{0.4t}$ to obtain $e^{0.4t}(0.35 + 0.14t - 1.68) = 0$ M1\n- Solves $0.14t - 1.33 = 0$ to find the exact value $t = 9.5$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_product_rule","label":"Direct Product Rule","steps":[{"type":"text","order":0,"title":"Identify the strategy","content":"To find the exact value of $t$ where the displacement $s$ has a stationary point, we need to:\n\n1. Find the derivative $\\frac{ds}{dt}$ using the **Product Rule**.\n2. Set $\\frac{ds}{dt} = 0$.\n3. Solve the resulting equation for $t$."},{"type":"text","order":1,"title":"Differentiate individual terms","content":"The displacement equation is given by:\n\n$$s = (0.35t - 4.2) e^{0.4t}$$\n\nThis is a product of two functions, $u(t)$ and $v(t)$:\n- Let $u = 0.35t - 4.2$, which differentiates to:\n$$\\frac{du}{dt} = 0.35$$\n\n- Let $v = e^{0.4t}$, which (using the chain rule) differentiates to:\n$$\\frac{dv}{dt} = 0.4e^{0.4t}$$\n\n*Note: This earns you the first mark B1.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Product Rule: } \\frac{ds}{dt} = \\htmlData{anchor=u-prime}{\\frac{du}{dt}}\\htmlData{anchor=v-term}{v} + \\htmlData{anchor=u-term}{u}\\htmlData{anchor=v-prime}{\\frac{dv}{dt}}$"},{"latex":"$$\\frac{ds}{dt} = \\htmlData{anchor=sub1}{\\color{@sky}{0.35}}\\htmlData{anchor=sub2}{\\color{@purple}{e^{0.4t}}} + \\htmlData{anchor=sub3}{\\color{@teal}{(0.35t - 4.2)}}\\htmlData{anchor=sub4}{\\color{@amber}{0.4e^{0.4t}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub1"},{"color":"purple","style":"text-color","anchor":"sub2"},{"color":"teal","style":"text-color","anchor":"sub3"},{"color":"amber","style":"text-color","anchor":"sub4"}],"connections":[{"to":"sub1","from":"u-prime","color":"sky","label":"u'","style":"solid"},{"to":"sub2","from":"v-term","color":"purple","label":"v","style":"solid"},{"to":"sub3","from":"u-term","color":"teal","label":"u","style":"solid"},{"to":"sub4","from":"v-prime","color":"amber","label":"v'","style":"solid"}]},"order":2,"title":"Apply the product rule","description":"We apply the Product Rule formula, substituting our components $u$, $v$, $u'$, and $v'$ to get the overall expression for the derivative. This earns the second mark M1."},{"type":"text","order":3,"title":"Factor out the exponential","content":"To locate the stationary points, we set the derivative to zero:\n\n$$0.35e^{0.4t} + 0.4(0.35t - 4.2)e^{0.4t} = 0$$\n\nSince $e^{0.4t}$ is a common factor in both terms, we can factorise it out:\n\n$$e^{0.4t} \\left[ 0.35 + 0.4(0.35t - 4.2) \\right] = 0$$\n\nNow, expand the terms inside the square bracket:\n\n$$e^{0.4t} (0.35 + 0.14t - 1.68) = 0$$\n$$e^{0.4t} (0.14t - 1.33) = 0$$\n\n*Factoring out the exponential term correctly earns you the third mark M1.*"},{"type":"text","order":4,"title":"Solve for t","content":"We have two factors multiplied together equaling zero:\n\n1. $e^{0.4t} = 0$\nSince the exponential function $e^x$ is always strictly positive ($e^{0.4t} > 0$ for all real $t$), this equation has **no real solutions**.\n\n2. $0.14t - 1.33 = 0$\nSolving this linear equation:\n\n$$\\begin{aligned} 0.14t &= 1.33 \\\\ t &= \\frac{1.33}{0.14} \\\\ t &= 9.5 \\end{aligned}$$\n\nThus, the exact value is **$t = 9.5$ seconds**.\n\n*This final calculation earns you the accuracy mark A1.*","calculator":[{"gdc":{"expression":"y = (0.35*x - 4.2)*e^(0.4*x)","instructions":"1. Go to Graph mode and enter the function: y = (0.35x - 4.2)e^(0.4x).\n2. Graph the function and use the G-Solv (F5) tool.\n3. Select MIN (F3) to find the coordinates of the local minimum."},"ti84":{"expression":"(0.35*X - 4.2) * e^(0.4*X)","instructions":"1. Press [Y=] and enter Y1 = (0.35X - 4.2) * e^(0.4X).\n2. Press [GRAPH] (adjust window if needed, e.g., Xmin = 0, Xmax = 15, Ymin = -15, Ymax = 10).\n3. Press [2nd] [CALC] and select 3: minimum (since this is a local minimum point).\n4. Set a Left Bound to the left of 9.5 (e.g., 8), a Right Bound to the right (e.g., 11), and press [ENTER] on Guess."},"desmos":{"expression":"y = (0.35*x - 4.2)*e^(0.4*x)","instructions":"1. Type the equation: y = (0.35x - 4.2)e^(0.4x).\n2. Click directly on the turning point (the local minimum) shown on the graph.\n3. The coordinates (9.5, -2.015) will appear, confirming $t = 9.5$."},"description":"Use your GDC to verify the solution by finding the root of the derivative or plotting the displacement function to find its minimum/maximum."}]}]},{"id":"expand_and_differentiate","label":"Expansion and Term-by-Term Differentiation","steps":[{"type":"text","order":0,"title":"Expand the equation","content":"To find the stationary point using the **Expansion and Term-by-Term Differentiation** method, we first expand the displacement expression to separate it into two distinct terms:\n\n$$s = (0.35t - 4.2) e^{0.4t}$$\n\nMultiplying $e^{0.4t}$ into the parentheses gives:\n\n$$s = 0.35t e^{0.4t} - 4.2 e^{0.4t}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$s = \\htmlData{anchor=term1}{\\color{@sky}{0.35t e^{0.4t}}} \\htmlData{anchor=term2}{\\color{@amber}{- 4.2 e^{0.4t}}}$"},{"latex":"$$\\frac{ds}{dt} = \\htmlData{anchor=diff1}{\\color{@sky}{\\left[0.35e^{0.4t} + 0.14t e^{0.4t}\\right]}} \\htmlData{anchor=diff2}{\\color{@amber}{- 1.68e^{0.4t}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term1"},{"color":"amber","style":"text-color","anchor":"term2"}],"connections":[{"to":"diff1","from":"term1","color":"sky","label":"Product Rule","style":"solid"},{"to":"diff2","from":"term2","color":"amber","label":"Chain Rule","style":"solid"}]},"order":1,"title":"Differentiate each term","description":"We now differentiate each term with respect to $t$ using the appropriate calculus rules."},{"type":"text","order":2,"title":"Details of differentiation","content":"Here is how the individual derivatives are obtained:\n\n1. **First Term:** For $0.35t e^{0.4t}$, we apply the product rule $\\frac{d}{dt}(uv) = u'v + uv'$ with $u = 0.35t$ and $v = e^{0.4t}$:\n $$\\frac{d}{dt}(0.35t e^{0.4t}) = 0.35e^{0.4t} + 0.35t(0.4e^{0.4t}) = 0.35e^{0.4t} + 0.14t e^{0.4t}$$\n\n2. **Second Term:** For $-4.2e^{0.4t}$, we apply the chain rule:\n $$\\frac{d}{dt}(-4.2e^{0.4t}) = -4.2 \\times 0.4 e^{0.4t} = -1.68e^{0.4t}$$\n\n*(These initial differentiation steps earn the first B1 and M1 marks)*"},{"type":"text","order":3,"title":"Combine and factorise","content":"Combine the differentiated terms together to get the full derivative:\n\n$$\\frac{ds}{dt} = 0.35e^{0.4t} + 0.14t e^{0.4t} - 1.68e^{0.4t}$$\n\nSince $e^{0.4t}$ is common to all terms, we can factorise it out:\n\n$$\\frac{ds}{dt} = e^{0.4t}(0.35 + 0.14t - 1.68)$$\n\nSimplify the terms inside the parentheses:\n\n$$\\frac{ds}{dt} = e^{0.4t}(0.14t - 1.33)$$"},{"type":"text","order":4,"title":"Solve for stationary point","content":"A stationary point occurs when the derivative is zero, so we set $\\frac{ds}{dt} = 0$:\n\n$$e^{0.4t}(0.14t - 1.33) = 0$$\n\n*(This step earns the M1 mark)*\n\nSince the exponential term $e^{0.4t}$ is always positive and can never equal zero for any real value of $t$, we can divide both sides by $e^{0.4t}$:\n\n$$0.14t - 1.33 = 0$$\n\nNow, solve for $t$:\n\n$$0.14t = 1.33$$\n\n$$t = \\frac{1.33}{0.14} = 9.5$$\n\nTherefore, the exact value of $t$ at the stationary point is $t = 9.5$ seconds.\n\n*(This final solution earns the A1 mark)*","calculator":[{"gdc":{"expression":"(0.35*x - 4.2) * e^(0.4*x)","instructions":"1. Go to the Graph menu and input the function f1(x) = (0.35x - 4.2) * e^(0.4x).\n2. Graph the function and use the 'Analyze Graph' -> 'Minimum' tool.\n3. Click on the left and right sides of the local minimum to find the coordinates of the turning point, which confirms the x-coordinate (time $t$) is 9.5."},"ti84":{"expression":"(0.35*X - 4.2)*e^(0.4*X)","instructions":"1. Press [Y=] and enter Y1 = (0.35*X - 4.2)*e^(0.4*X).\n2. Press [GRAPH] and adjust the window to see the turning point (e.g., Xmin = 0, Xmax = 15, Ymin = -10, Ymax = 10).\n3. Press [2nd] [TRACE] and select '3: minimum'.\n4. Set a Left Bound to the left of the turning point, a Right Bound to the right, and press [ENTER] on Guess to find the stationary point at X = 9.5."},"desmos":{"expression":"y = (0.35x - 4.2) * e^(0.4x)","instructions":"1. In the first input bar, type: y = (0.35x - 4.2) * e^(0.4x).\n2. Click on the gray point representing the local minimum on the graph.\n3. The label will display (9.5, -2.052), showing that the stationary point occurs exactly at $t = 9.5$."},"description":"We can verify the solution on a graphing utility by plotting the derivative function or finding the minimum/maximum of the original displacement function."}]}]},{"id":"logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Introduce absolute values","content":"To find the stationary points, we need to differentiate the displacement function:\n\n$$s = (0.35t - 4.2) e^{0.4t}$$\n\nSince the term $(0.35t - 4.2)$ can be negative, we use the absolute value of both sides before taking the natural logarithm. This ensures that the logarithm is mathematically defined for real numbers:\n\n$$|s| = |0.35t - 4.2| \\cdot e^{0.4t}$$"},{"type":"text","order":1,"title":"Apply natural logarithm","content":"Now we take the natural logarithm ($\\ln$) of both sides. Using the logarithmic product property $\\ln(ab) = \\ln(a) + \\ln(b)$ and power property $\\ln(e^x) = x$, we convert the multiplication into addition:\n\n$$\\ln|s| = \\ln|0.35t - 4.2| + \\ln\\left(e^{0.4t}\\right)$$\n\n$$\\ln|s| = \\ln|0.35t - 4.2| + 0.4t$$"},{"type":"text","order":2,"title":"Differentiate implicitly","content":"Next, we differentiate both sides with respect to $t$. \n\nRecall that the derivative of $\\ln|f(x)|$ with respect to $x$ is simply $\\frac{f'(x)}{f(x)}$:\n\n$$\\frac{1}{s}\\frac{ds}{dt} = \\frac{0.35}{0.35t - 4.2} + 0.4$$"},{"type":"text","order":3,"title":"Set derivative to zero","content":"At a stationary point, the rate of change of displacement is zero, so we set $\\frac{ds}{dt} = 0$:\n\n$$\\frac{1}{s}(0) = \\frac{0.35}{0.35t - 4.2} + 0.4$$\n\n$$0 = \\frac{0.35}{0.35t - 4.2} + 0.4$$"},{"type":"text","order":4,"title":"Solve for t","content":"We now solve this equation for $t$:\n\n$$-0.4 = \\frac{0.35}{0.35t - 4.2}$$\n\nMultiply both sides by $(0.35t - 4.2)$:\n\n$$-0.4(0.35t - 4.2) = 0.35$$\n\n$$-0.14t + 1.68 = 0.35$$\n\n$$-0.14t = 0.35 - 1.68$$\n\n$$-0.14t = -1.33$$\n\n$$t = \\frac{-1.33}{-0.14} = 9.5$$"},{"type":"text","order":5,"title":"State final answer","content":"The exact value of $t$ at the stationary point is:\n\n$$t = 9.5 \\text{ seconds}$$"}]}],"generatedAt":"2026-06-09T07:05:14.631Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"f196addb-e240-4981-9709-25479d539507:direct_product_rule","label":"Direct Product Rule","steps":[{"type":"text","order":0,"title":"Identify the strategy","content":"To find the exact value of $t$ where the displacement $s$ has a stationary point, we need to:\n\n1. Find the derivative $\\frac{ds}{dt}$ using the **Product Rule**.\n2. Set $\\frac{ds}{dt} = 0$.\n3. Solve the resulting equation for $t$."},{"type":"text","order":1,"title":"Differentiate individual terms","content":"The displacement equation is given by:\n\n$$s = (0.35t - 4.2) e^{0.4t}$$\n\nThis is a product of two functions, $u(t)$ and $v(t)$:\n- Let $u = 0.35t - 4.2$, which differentiates to:\n$$\\frac{du}{dt} = 0.35$$\n\n- Let $v = e^{0.4t}$, which (using the chain rule) differentiates to:\n$$\\frac{dv}{dt} = 0.4e^{0.4t}$$\n\n*Note: This earns you the first mark B1.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Product Rule: } \\frac{ds}{dt} = \\htmlData{anchor=u-prime}{\\frac{du}{dt}}\\htmlData{anchor=v-term}{v} + \\htmlData{anchor=u-term}{u}\\htmlData{anchor=v-prime}{\\frac{dv}{dt}}$"},{"latex":"$$\\frac{ds}{dt} = \\htmlData{anchor=sub1}{\\color{@sky}{0.35}}\\htmlData{anchor=sub2}{\\color{@purple}{e^{0.4t}}} + \\htmlData{anchor=sub3}{\\color{@teal}{(0.35t - 4.2)}}\\htmlData{anchor=sub4}{\\color{@amber}{0.4e^{0.4t}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub1"},{"color":"purple","style":"text-color","anchor":"sub2"},{"color":"teal","style":"text-color","anchor":"sub3"},{"color":"amber","style":"text-color","anchor":"sub4"}],"connections":[{"to":"sub1","from":"u-prime","color":"sky","label":"u'","style":"solid"},{"to":"sub2","from":"v-term","color":"purple","label":"v","style":"solid"},{"to":"sub3","from":"u-term","color":"teal","label":"u","style":"solid"},{"to":"sub4","from":"v-prime","color":"amber","label":"v'","style":"solid"}]},"order":2,"title":"Apply the product rule","description":"We apply the Product Rule formula, substituting our components $u$, $v$, $u'$, and $v'$ to get the overall expression for the derivative. This earns the second mark M1."},{"type":"text","order":3,"title":"Factor out the exponential","content":"To locate the stationary points, we set the derivative to zero:\n\n$$0.35e^{0.4t} + 0.4(0.35t - 4.2)e^{0.4t} = 0$$\n\nSince $e^{0.4t}$ is a common factor in both terms, we can factorise it out:\n\n$$e^{0.4t} \\left[ 0.35 + 0.4(0.35t - 4.2) \\right] = 0$$\n\nNow, expand the terms inside the square bracket:\n\n$$e^{0.4t} (0.35 + 0.14t - 1.68) = 0$$\n$$e^{0.4t} (0.14t - 1.33) = 0$$\n\n*Factoring out the exponential term correctly earns you the third mark M1.*"},{"type":"text","order":4,"title":"Solve for t","content":"We have two factors multiplied together equaling zero:\n\n1. $e^{0.4t} = 0$\nSince the exponential function $e^x$ is always strictly positive ($e^{0.4t} > 0$ for all real $t$), this equation has **no real solutions**.\n\n2. $0.14t - 1.33 = 0$\nSolving this linear equation:\n\n$$\\begin{aligned} 0.14t &= 1.33 \\\\ t &= \\frac{1.33}{0.14} \\\\ t &= 9.5 \\end{aligned}$$\n\nThus, the exact value is **$t = 9.5$ seconds**.\n\n*This final calculation earns you the accuracy mark A1.*","calculator":[{"gdc":{"expression":"y = (0.35*x - 4.2)*e^(0.4*x)","instructions":"1. Go to Graph mode and enter the function: y = (0.35x - 4.2)e^(0.4x).\n2. Graph the function and use the G-Solv (F5) tool.\n3. Select MIN (F3) to find the coordinates of the local minimum."},"ti84":{"expression":"(0.35*X - 4.2) * e^(0.4*X)","instructions":"1. Press [Y=] and enter Y1 = (0.35X - 4.2) * e^(0.4X).\n2. Press [GRAPH] (adjust window if needed, e.g., Xmin = 0, Xmax = 15, Ymin = -15, Ymax = 10).\n3. Press [2nd] [CALC] and select 3: minimum (since this is a local minimum point).\n4. Set a Left Bound to the left of 9.5 (e.g., 8), a Right Bound to the right (e.g., 11), and press [ENTER] on Guess."},"desmos":{"expression":"y = (0.35*x - 4.2)*e^(0.4*x)","instructions":"1. Type the equation: y = (0.35x - 4.2)e^(0.4x).\n2. Click directly on the turning point (the local minimum) shown on the graph.\n3. The coordinates (9.5, -2.015) will appear, confirming $t = 9.5$."},"description":"Use your GDC to verify the solution by finding the root of the derivative or plotting the displacement function to find its minimum/maximum."}]}]},{"id":"f196addb-e240-4981-9709-25479d539507:expand_and_differentiate","label":"Expansion and Term-by-Term Differentiation","steps":[{"type":"text","order":0,"title":"Expand the equation","content":"To find the stationary point using the **Expansion and Term-by-Term Differentiation** method, we first expand the displacement expression to separate it into two distinct terms:\n\n$$s = (0.35t - 4.2) e^{0.4t}$$\n\nMultiplying $e^{0.4t}$ into the parentheses gives:\n\n$$s = 0.35t e^{0.4t} - 4.2 e^{0.4t}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$s = \\htmlData{anchor=term1}{\\color{@sky}{0.35t e^{0.4t}}} \\htmlData{anchor=term2}{\\color{@amber}{- 4.2 e^{0.4t}}}$"},{"latex":"$$\\frac{ds}{dt} = \\htmlData{anchor=diff1}{\\color{@sky}{\\left[0.35e^{0.4t} + 0.14t e^{0.4t}\\right]}} \\htmlData{anchor=diff2}{\\color{@amber}{- 1.68e^{0.4t}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term1"},{"color":"amber","style":"text-color","anchor":"term2"}],"connections":[{"to":"diff1","from":"term1","color":"sky","label":"Product Rule","style":"solid"},{"to":"diff2","from":"term2","color":"amber","label":"Chain Rule","style":"solid"}]},"order":1,"title":"Differentiate each term","description":"We now differentiate each term with respect to $t$ using the appropriate calculus rules."},{"type":"text","order":2,"title":"Details of differentiation","content":"Here is how the individual derivatives are obtained:\n\n1. **First Term:** For $0.35t e^{0.4t}$, we apply the product rule $\\frac{d}{dt}(uv) = u'v + uv'$ with $u = 0.35t$ and $v = e^{0.4t}$:\n $$\\frac{d}{dt}(0.35t e^{0.4t}) = 0.35e^{0.4t} + 0.35t(0.4e^{0.4t}) = 0.35e^{0.4t} + 0.14t e^{0.4t}$$\n\n2. **Second Term:** For $-4.2e^{0.4t}$, we apply the chain rule:\n $$\\frac{d}{dt}(-4.2e^{0.4t}) = -4.2 \\times 0.4 e^{0.4t} = -1.68e^{0.4t}$$\n\n*(These initial differentiation steps earn the first B1 and M1 marks)*"},{"type":"text","order":3,"title":"Combine and factorise","content":"Combine the differentiated terms together to get the full derivative:\n\n$$\\frac{ds}{dt} = 0.35e^{0.4t} + 0.14t e^{0.4t} - 1.68e^{0.4t}$$\n\nSince $e^{0.4t}$ is common to all terms, we can factorise it out:\n\n$$\\frac{ds}{dt} = e^{0.4t}(0.35 + 0.14t - 1.68)$$\n\nSimplify the terms inside the parentheses:\n\n$$\\frac{ds}{dt} = e^{0.4t}(0.14t - 1.33)$$"},{"type":"text","order":4,"title":"Solve for stationary point","content":"A stationary point occurs when the derivative is zero, so we set $\\frac{ds}{dt} = 0$:\n\n$$e^{0.4t}(0.14t - 1.33) = 0$$\n\n*(This step earns the M1 mark)*\n\nSince the exponential term $e^{0.4t}$ is always positive and can never equal zero for any real value of $t$, we can divide both sides by $e^{0.4t}$:\n\n$$0.14t - 1.33 = 0$$\n\nNow, solve for $t$:\n\n$$0.14t = 1.33$$\n\n$$t = \\frac{1.33}{0.14} = 9.5$$\n\nTherefore, the exact value of $t$ at the stationary point is $t = 9.5$ seconds.\n\n*(This final solution earns the A1 mark)*","calculator":[{"gdc":{"expression":"(0.35*x - 4.2) * e^(0.4*x)","instructions":"1. Go to the Graph menu and input the function f1(x) = (0.35x - 4.2) * e^(0.4x).\n2. Graph the function and use the 'Analyze Graph' -> 'Minimum' tool.\n3. Click on the left and right sides of the local minimum to find the coordinates of the turning point, which confirms the x-coordinate (time $t$) is 9.5."},"ti84":{"expression":"(0.35*X - 4.2)*e^(0.4*X)","instructions":"1. Press [Y=] and enter Y1 = (0.35*X - 4.2)*e^(0.4*X).\n2. Press [GRAPH] and adjust the window to see the turning point (e.g., Xmin = 0, Xmax = 15, Ymin = -10, Ymax = 10).\n3. Press [2nd] [TRACE] and select '3: minimum'.\n4. Set a Left Bound to the left of the turning point, a Right Bound to the right, and press [ENTER] on Guess to find the stationary point at X = 9.5."},"desmos":{"expression":"y = (0.35x - 4.2) * e^(0.4x)","instructions":"1. In the first input bar, type: y = (0.35x - 4.2) * e^(0.4x).\n2. Click on the gray point representing the local minimum on the graph.\n3. The label will display (9.5, -2.052), showing that the stationary point occurs exactly at $t = 9.5$."},"description":"We can verify the solution on a graphing utility by plotting the derivative function or finding the minimum/maximum of the original displacement function."}]}]},{"id":"f196addb-e240-4981-9709-25479d539507:logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Introduce absolute values","content":"To find the stationary points, we need to differentiate the displacement function:\n\n$$s = (0.35t - 4.2) e^{0.4t}$$\n\nSince the term $(0.35t - 4.2)$ can be negative, we use the absolute value of both sides before taking the natural logarithm. This ensures that the logarithm is mathematically defined for real numbers:\n\n$$|s| = |0.35t - 4.2| \\cdot e^{0.4t}$$"},{"type":"text","order":1,"title":"Apply natural logarithm","content":"Now we take the natural logarithm ($\\ln$) of both sides. Using the logarithmic product property $\\ln(ab) = \\ln(a) + \\ln(b)$ and power property $\\ln(e^x) = x$, we convert the multiplication into addition:\n\n$$\\ln|s| = \\ln|0.35t - 4.2| + \\ln\\left(e^{0.4t}\\right)$$\n\n$$\\ln|s| = \\ln|0.35t - 4.2| + 0.4t$$"},{"type":"text","order":2,"title":"Differentiate implicitly","content":"Next, we differentiate both sides with respect to $t$. \n\nRecall that the derivative of $\\ln|f(x)|$ with respect to $x$ is simply $\\frac{f'(x)}{f(x)}$:\n\n$$\\frac{1}{s}\\frac{ds}{dt} = \\frac{0.35}{0.35t - 4.2} + 0.4$$"},{"type":"text","order":3,"title":"Set derivative to zero","content":"At a stationary point, the rate of change of displacement is zero, so we set $\\frac{ds}{dt} = 0$:\n\n$$\\frac{1}{s}(0) = \\frac{0.35}{0.35t - 4.2} + 0.4$$\n\n$$0 = \\frac{0.35}{0.35t - 4.2} + 0.4$$"},{"type":"text","order":4,"title":"Solve for t","content":"We now solve this equation for $t$:\n\n$$-0.4 = \\frac{0.35}{0.35t - 4.2}$$\n\nMultiply both sides by $(0.35t - 4.2)$:\n\n$$-0.4(0.35t - 4.2) = 0.35$$\n\n$$-0.14t + 1.68 = 0.35$$\n\n$$-0.14t = 0.35 - 1.68$$\n\n$$-0.14t = -1.33$$\n\n$$t = \\frac{-1.33}{-0.14} = 9.5$$"},{"type":"text","order":5,"title":"State final answer","content":"The exact value of $t$ at the stationary point is:\n\n$$t = 9.5 \\text{ seconds}$$"}]}],"generatedAt":"2026-06-09T07:05:27.437Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"e0b9b73d-7217-4a30-9f07-a62b43ccabce","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"8423de37-6e55-4383-8387-71cd38a57705","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"e10def56-640b-416a-b974-27935fa57ad3","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"ed0d3408-c4dc-41f2-9018-0ac8079562aa","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"f2d4c50a-dfb8-4a87-a753-18a9f23a84cc","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"66f47b42-061b-447f-af98-b82c19ee318b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"c3f9a0ad-9898-4b93-8e8f-6da9a3e5e6b5","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"6c577d2b-91b1-48d8-b0d0-44c32c8af6a3","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"9ec7b6f4-b03d-4589-a8f9-3cce13a23630","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"b688a254-1710-4fae-ba71-47995d3640d7","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"f19c6214-3584-4bc8-8bf3-48d78b7096e3","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"68dcddf7-2ec6-4759-ae74-f44fdd72209a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"60601d3d-8f39-40ad-a875-04eae3ad5f41","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"79329219-0030-4556-9154-62cbbc920628","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"aece7605-9257-44ce-9d55-4f97e298e9e9","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"b9cb05cf-d731-4962-afed-d69e3dacbe8f","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"3afcef57-5d5d-4b3d-a83f-6139c6a94835","difficultyBand":"medium","calculatorUse":null,"totalMarks":14,"questionType":"LA","isCalculation":true},{"id":"459f1ce7-63d8-41b2-b571-c050d1bef394","difficultyBand":"hard","calculatorUse":null,"totalMarks":14,"questionType":"LA","isCalculation":true},{"id":"628b555b-bb68-4df3-9cd9-fee55098c4c7","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"652a6a94-778e-4806-83a1-f819e484bd63","difficultyBand":"medium","calculatorUse":null,"totalMarks":13,"questionType":"LA","isCalculation":true},{"id":"763347ec-121f-4b55-9b63-860c6d3642f6","difficultyBand":"medium","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"798c117d-a4ff-46de-9f87-692828cce780","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"8ebc1394-5cdb-455f-a25d-6766c150b7ba","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"a1d2b6ec-df5d-4f7b-8c86-c5b92bb30c67","difficultyBand":"medium","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"b0d0eb60-ceb7-491d-9ba2-818dd9ecc01f","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"c0383345-f45a-48b6-851c-44384d7b70af","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"e6789c8e-b45a-4317-ad00-f16d2eb6fef6","difficultyBand":"medium","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"e79c5d66-d3a4-4787-9db4-a8bda6f14225","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"fa7bf159-3159-46a7-8020-178fbdeb932f","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"fa9f5fd2-5fe2-4b80-a97b-04003fe5fa85","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true}],"title":"9.4 The Product Rule","topicId":"655d0a31-774e-40b3-985a-a19d7c6e3f02","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-9-4-the-product-rule","topicTitle":"9.4 The Product Rule"}]}],"$L3c","$L3d"]}]

9.4 The Product Rule

Question 7

9.4 The Product Rule Questions

9.4 The Product Rule

Question 7

9.4 The Product Rule Questions