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expression $\\frac{dy}{dx} = \\frac{2 - 6x - y}{x + 4y}$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Differentiates $3x^2 - 2x$ to $6x - 2$ and $2y^2$ to $4y \\frac{dy}{dx}$ accurately","code":"M"},{"id":"p2","marks":1,"description":"Differentiates $xy$ using product rule correctly","code":"M"},{"id":"p3","marks":1,"description":"Correctly handles derivative of the constant 11","code":"B"},{"id":"p4","marks":1,"description":"Collects $\\frac{dy}{dx}$ terms and factorises","code":"M"},{"id":"p5","marks":1,"description":"Final correct formula for $\\frac{dy}{dx}$","code":"A"}],"finalAnswers":["\\frac{dy}{dx} = \\frac{2 - 6x - y}{x + 4y}","\\frac{dy}{dx} = -\\frac{6x + y - 2}{x + 4y}"],"generatedAt":"2026-05-06T14:04:27.582Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"implicit_differentiation","label":"Implicit Differentiation","steps":[{"type":"text","order":0,"title":"Identify the approach","content":"To find $\\frac{dy}{dx}$, we will use **implicit differentiation**. This involves differentiating both sides of the equation term-by-term with respect to $x$, applying the **chain rule** and **product rule** where necessary."},{"type":"text","order":1,"title":"Differentiate simple terms","content":"Let's first differentiate the terms that depend only on $x$, as well as the constant on the right-hand side:\n\n* The derivative of $3x^2$ with respect to $x$ is $6x$.\n* The derivative of $-2x$ with respect to $x$ is $-2$.\n* The derivative of the constant $11$ with respect to $x$ is $0$.\n\nThis earns us our first mark (**B1** for the right-hand side constant derivative)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}\\left(\\htmlData{anchor=y-term}{\\color{@sky}{2y^2}}\\right) = \\frac{d}{dy}\\left(\\color{@sky}{2y^2}\\right) \\cdot \\htmlData{anchor=chain-dy-dx}{\\color{@amber}{\\frac{dy}{dx}}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(2y^2\\right)} = \\htmlData{anchor=deriv-y}{\\color{@sky}{4y}} \\htmlData{anchor=dy-dx}{\\color{@amber}{\\frac{dy}{dx}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-term"},{"color":"amber","style":"text-color","anchor":"chain-dy-dx"},{"color":"sky","style":"text-color","anchor":"deriv-y"},{"color":"amber","style":"text-color","anchor":"dy-dx"}],"connections":[{"to":"dy-dx","from":"chain-dy-dx","color":"amber","label":"chain rule term","style":"solid"}]},"order":2,"title":"Differentiate the y-term","description":"For the term involving $2y^2$, we must apply the chain rule. Differentiate with respect to $y$ and multiply by $\\frac{dy}{dx}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}\\left(\\htmlData{anchor=u-term}{\\color{@sky}{x}}\\htmlData{anchor=v-term}{\\color{@purple}{y}}\\right) = \\htmlData{anchor=v-term-diff}{\\color{@purple}{y}}\\frac{d}{dx}(\\color{@sky}{x}) + \\color{@sky}{x}\\frac{d}{dx}(\\htmlData{anchor=y-diff}{\\color{@purple}{y}})$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(xy\\right)} = \\color{@purple}{y}(1) + \\color{@sky}{x}\\left(\\htmlData{anchor=dy-dx-prod}{\\color{@amber}{\\frac{dy}{dx}}}\\right)$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(xy\\right)} = \\htmlData{anchor=final-prod-diff}{\\color{@purple}{y} + \\color{@sky}{x}\\color{@amber}{\\frac{dy}{dx}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"purple","style":"text-color","anchor":"v-term"},{"color":"teal","style":"box","anchor":"final-prod-diff"}],"connections":[{"to":"dy-dx-prod","from":"y-diff","color":"amber","label":"derivative with respect to x","style":"solid"}]},"order":3,"title":"Differentiate the product xy","description":"The term $xy$ is a product of $x$ and $y$. We must use the product rule: $\\frac{d}{dx}(uv) = v \\frac{du}{dx} + u \\frac{dv}{dx}$."},{"type":"text","order":4,"title":"Combine differentiated terms","content":"Now we assemble all of our differentiated terms into a single equation:\n\n$$(6x - 2) + \\left(y + x \\frac{dy}{dx}\\right) + 4y \\frac{dy}{dx} = 0$$\n\nThis earns the first two method marks (**M1**, **M1**) for correctly differentiating each term."},{"type":"text","order":5,"title":"Rearrange and factorise","content":"To solve for $\\frac{dy}{dx}$, first group all terms containing $\\frac{dy}{dx}$ on one side, and move the remaining terms to the other side:\n\n$$x \\frac{dy}{dx} + 4y \\frac{dy}{dx} = 2 - 6x - y$$\n\nNow, factorise out $\\frac{dy}{dx}$ on the left-hand side (**M1**):\n\n$$\\frac{dy}{dx} (x + 4y) = 2 - 6x - y$$"},{"type":"text","order":6,"title":"Isolate the derivative","content":"Finally, divide both sides by $(x + 4y)$ to isolate $\\frac{dy}{dx}$ (**A1**):\n\n$$\\frac{dy}{dx} = \\frac{2 - 6x - y}{x + 4y}$$"}]}],"generatedAt":"2026-06-09T01:59:36.853Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"763d3587-a147-43c9-a9b8-8eed0d9c1ccd","label":"b","content":"Find an equation for the tangent to the curve at the point $P(1, 2)$","markscheme":"- Substitutes $(1, 2)$ into their $\\frac{dy}{dx}$ to find gradient $-\\frac{2}{3}$ M1\n- Uses $y - y_1 = m(x - x_1)$ with their gradient M1\n- Correct equation $y = -\\frac{2}{3}x + \\frac{8}{3}$ or $2x + 3y = 8$ or equivalent A1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Calculates gradient at $(1, 2)$ using their expression","code":"M"},{"id":"p2","marks":1,"description":"Applies the point-slope form with $P(1, 2)$ and their $m$","code":"M"},{"id":"p3","marks":1,"description":"Simplifies to a correct tangent equation","code":"A"}],"finalAnswers":["y = -\\frac{2}{3}x + \\frac{8}{3}","2x + 3y = 8","y - 2 = -\\frac{2}{3}(x - 1)"],"generatedAt":"2026-05-06T14:04:27.582Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"implicit_differentiation","label":"Implicit Differentiation","steps":[{"type":"text","order":0,"title":"Differentiate implicitly","content":"To find the gradient of the curve, we differentiate the equation $3x^2 + xy - 2x + 2y^2 = 11$ with respect to $x$.\n\nRemember to apply:\n\n* The **product rule** on the $xy$ term: $\\frac{d}{dx}(xy) = x\\frac{dy}{dx} + y$\n* The **chain rule** on the $2y^2$ term: $\\frac{d}{dx}(2y^2) = 4y\\frac{dy}{dx}$\n\nDifferentiating term by term gives:\n\n$$6x + \\left(x\\frac{dy}{dx} + y\\right) - 2 + 4y\\frac{dy}{dx} = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$6x + y - 2 + \\htmlData{anchor=dy-terms}{\\color{@sky}{\\left(x + 4y\\right)\\frac{dy}{dx}}} = 0$"},{"latex":"$$\\htmlData{anchor=dy-iso}{\\color{@sky}{\\frac{dy}{dx}}} = \\frac{\\htmlData{anchor=rhs}{\\color{@amber}{2 - 6x - y}}}{\\htmlData{anchor=denom}{\\color{@pink}{x + 4y}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dy-terms"},{"color":"sky","style":"text-color","anchor":"dy-iso"}],"connections":[{"to":"dy-iso","from":"dy-terms","color":"sky","label":"isolate dy/dx","style":"solid"}]},"order":1,"title":"Solve for the derivative","description":"Rearranging the equation to solve for $\\frac{dy}{dx}$ allows us to easily compute the gradient at any given point."},{"type":"text","order":2,"title":"Find the gradient","content":"Now we find the gradient $m$ of the tangent at the point $P(1, 2)$ by substituting $x = 1$ and $y = 2$ into our expression for $\\frac{dy}{dx}$:\n\n$$m = \\frac{2 - 6(1) - 2}{1 + 4(2)}$$\n\nCalculating this gives:\n\n$$m = \\frac{-6}{9} = -\\frac{2}{3}$$","highlights":[{"text":"P(1, 2)","location":"specification"}]},{"type":"text","order":3,"title":"Find tangent line equation","content":"With the gradient $m = -\\frac{2}{3}$ and the point $P(1, 2)$, we use the point-gradient formula $y - y_1 = m(x - x_1)$:\n\n$$y - 2 = -\\frac{2}{3}(x - 1)$$\n\nMultiplying through by $3$ to remove fractions:\n\n$$3(y - 2) = -2(x - 1)$$\n\n$$3y - 6 = -2x + 2$$\n\nRearranging into standard form gives:\n\n$$2x + 3y = 8$$"}]},{"id":"conic_tangent_formula","label":"Conic Tangent Formula","steps":[{"type":"text","order":0,"title":"Introduce the Tangent Rule","content":"For any second-degree conic curve, we can find the equation of the tangent at a point $(x_1, y_1)$ on the curve directly. We do this by replacing the quadratic and linear terms using the standard **$T = 0$ linear substitution rules**:\n\n* $x^2 \\to x_1x$\n* $y^2 \\to y_1y$\n* $xy \\to \\frac{x_1y + y_1x}{2}$\n* $x \\to \\frac{x + x_1}{2}$\n* $y \\to \\frac{y + y_1}{2}$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Curve equation:"},{"latex":"$$3\\htmlData{anchor=x2}{x^2} + \\htmlData{anchor=xy}{xy} - 2\\htmlData{anchor=x}{x} + 2\\htmlData{anchor=y2}{y^2} = 11$"},{"text":"Tangent equation:"},{"latex":"$$3\\htmlData{anchor=x1x}{(x_1x)} + \\htmlData{anchor=xysub}{\\frac{x_1y + y_1x}{2}} - 2\\htmlData{anchor=xsub}{\\left(\\frac{x + x_1}{2}\\right)} + 2\\htmlData{anchor=y1y}{(y_1y)} = 11$"}],"highlights":[{"color":"sky","style":"underline","anchor":"x2"},{"color":"amber","style":"underline","anchor":"xy"},{"color":"purple","style":"underline","anchor":"x"},{"color":"green","style":"underline","anchor":"y2"}],"connections":[{"to":"x1x","from":"x2","color":"sky","label":null,"style":"solid"},{"to":"xysub","from":"xy","color":"amber","label":null,"style":"solid"},{"to":"xsub","from":"x","color":"purple","label":null,"style":"solid"},{"to":"y1y","from":"y2","color":"green","label":null,"style":"solid"}]},"order":1,"title":"Apply substitution rules","description":"We replace each term of our curve equation $3x^2 + xy - 2x + 2y^2 = 11$ with its corresponding tangent substitution."},{"type":"text","order":2,"title":"Substitute the point P(1, 2)","content":"Now, we substitute the coordinates of the point $P(x_1, y_1) = (1, 2)$ into our tangent equation:\n\n$$3(1)x + \\frac{(1)y + (2)x}{2} - 2\\left(\\frac{x + 1}{2}\\right) + 2(2)y = 11$$\n\nSimplifying this term by term:\n\n$$3x + \\frac{y + 2x}{2} - (x + 1) + 4y = 11$$"},{"type":"text","order":3,"title":"Simplify and solve","content":"To eliminate the fraction, multiply the entire equation by $2$:\n\n$$6x + (y + 2x) - 2(x + 1) + 8y = 22$$\n\nExpand and collect like terms:\n\n$$\\begin{aligned} 6x + y + 2x - 2x - 2 + 8y &= 22 \\\\ 6x + 9y - 2 &= 22 \\\\ 6x + 9y &= 24 \\end{aligned}$$\n\nDividing the entire equation by $3$ gives the simplified linear equation:\n\n$$2x + 3y = 8$$"}]}],"generatedAt":"2026-06-09T02:01:29.711Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"5e82e939-6776-447c-9842-62b4a9b3b28c:implicit_differentiation","label":"(a) Implicit Differentiation","steps":[{"type":"text","order":0,"title":"Identify the approach","content":"To find $\\frac{dy}{dx}$, we will use **implicit differentiation**. This involves differentiating both sides of the equation term-by-term with respect to $x$, applying the **chain rule** and **product rule** where necessary."},{"type":"text","order":1,"title":"Differentiate simple terms","content":"Let's first differentiate the terms that depend only on $x$, as well as the constant on the right-hand side:\n\n* The derivative of $3x^2$ with respect to $x$ is $6x$.\n* The derivative of $-2x$ with respect to $x$ is $-2$.\n* The derivative of the constant $11$ with respect to $x$ is $0$.\n\nThis earns us our first mark (**B1** for the right-hand side constant derivative)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}\\left(\\htmlData{anchor=y-term}{\\color{@sky}{2y^2}}\\right) = \\frac{d}{dy}\\left(\\color{@sky}{2y^2}\\right) \\cdot \\htmlData{anchor=chain-dy-dx}{\\color{@amber}{\\frac{dy}{dx}}}$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(2y^2\\right)} = \\htmlData{anchor=deriv-y}{\\color{@sky}{4y}} \\htmlData{anchor=dy-dx}{\\color{@amber}{\\frac{dy}{dx}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-term"},{"color":"amber","style":"text-color","anchor":"chain-dy-dx"},{"color":"sky","style":"text-color","anchor":"deriv-y"},{"color":"amber","style":"text-color","anchor":"dy-dx"}],"connections":[{"to":"dy-dx","from":"chain-dy-dx","color":"amber","label":"chain rule term","style":"solid"}]},"order":2,"title":"Differentiate the y-term","description":"For the term involving $2y^2$, we must apply the chain rule. Differentiate with respect to $y$ and multiply by $\\frac{dy}{dx}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{d}{dx}\\left(\\htmlData{anchor=u-term}{\\color{@sky}{x}}\\htmlData{anchor=v-term}{\\color{@purple}{y}}\\right) = \\htmlData{anchor=v-term-diff}{\\color{@purple}{y}}\\frac{d}{dx}(\\color{@sky}{x}) + \\color{@sky}{x}\\frac{d}{dx}(\\htmlData{anchor=y-diff}{\\color{@purple}{y}})$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(xy\\right)} = \\color{@purple}{y}(1) + \\color{@sky}{x}\\left(\\htmlData{anchor=dy-dx-prod}{\\color{@amber}{\\frac{dy}{dx}}}\\right)$"},{"latex":"$$\\phantom{\\frac{d}{dx}\\left(xy\\right)} = \\htmlData{anchor=final-prod-diff}{\\color{@purple}{y} + \\color{@sky}{x}\\color{@amber}{\\frac{dy}{dx}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"purple","style":"text-color","anchor":"v-term"},{"color":"teal","style":"box","anchor":"final-prod-diff"}],"connections":[{"to":"dy-dx-prod","from":"y-diff","color":"amber","label":"derivative with respect to x","style":"solid"}]},"order":3,"title":"Differentiate the product xy","description":"The term $xy$ is a product of $x$ and $y$. We must use the product rule: $\\frac{d}{dx}(uv) = v \\frac{du}{dx} + u \\frac{dv}{dx}$."},{"type":"text","order":4,"title":"Combine differentiated terms","content":"Now we assemble all of our differentiated terms into a single equation:\n\n$$(6x - 2) + \\left(y + x \\frac{dy}{dx}\\right) + 4y \\frac{dy}{dx} = 0$$\n\nThis earns the first two method marks (**M1**, **M1**) for correctly differentiating each term."},{"type":"text","order":5,"title":"Rearrange and factorise","content":"To solve for $\\frac{dy}{dx}$, first group all terms containing $\\frac{dy}{dx}$ on one side, and move the remaining terms to the other side:\n\n$$x \\frac{dy}{dx} + 4y \\frac{dy}{dx} = 2 - 6x - y$$\n\nNow, factorise out $\\frac{dy}{dx}$ on the left-hand side (**M1**):\n\n$$\\frac{dy}{dx} (x + 4y) = 2 - 6x - y$$"},{"type":"text","order":6,"title":"Isolate the derivative","content":"Finally, divide both sides by $(x + 4y)$ to isolate $\\frac{dy}{dx}$ (**A1**):\n\n$$\\frac{dy}{dx} = \\frac{2 - 6x - y}{x + 4y}$$"}]},{"id":"763d3587-a147-43c9-a9b8-8eed0d9c1ccd:implicit_differentiation","label":"(b) Implicit Differentiation","steps":[{"type":"text","order":0,"title":"Differentiate implicitly","content":"To find the gradient of the curve, we differentiate the equation $3x^2 + xy - 2x + 2y^2 = 11$ with respect to $x$.\n\nRemember to apply:\n\n* The **product rule** on the $xy$ term: $\\frac{d}{dx}(xy) = x\\frac{dy}{dx} + y$\n* The **chain rule** on the $2y^2$ term: $\\frac{d}{dx}(2y^2) = 4y\\frac{dy}{dx}$\n\nDifferentiating term by term gives:\n\n$$6x + \\left(x\\frac{dy}{dx} + y\\right) - 2 + 4y\\frac{dy}{dx} = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$6x + y - 2 + \\htmlData{anchor=dy-terms}{\\color{@sky}{\\left(x + 4y\\right)\\frac{dy}{dx}}} = 0$"},{"latex":"$$\\htmlData{anchor=dy-iso}{\\color{@sky}{\\frac{dy}{dx}}} = \\frac{\\htmlData{anchor=rhs}{\\color{@amber}{2 - 6x - y}}}{\\htmlData{anchor=denom}{\\color{@pink}{x + 4y}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dy-terms"},{"color":"sky","style":"text-color","anchor":"dy-iso"}],"connections":[{"to":"dy-iso","from":"dy-terms","color":"sky","label":"isolate dy/dx","style":"solid"}]},"order":1,"title":"Solve for the derivative","description":"Rearranging the equation to solve for $\\frac{dy}{dx}$ allows us to easily compute the gradient at any given point."},{"type":"text","order":2,"title":"Find the gradient","content":"Now we find the gradient $m$ of the tangent at the point $P(1, 2)$ by substituting $x = 1$ and $y = 2$ into our expression for $\\frac{dy}{dx}$:\n\n$$m = \\frac{2 - 6(1) - 2}{1 + 4(2)}$$\n\nCalculating this gives:\n\n$$m = \\frac{-6}{9} = -\\frac{2}{3}$$","highlights":[{"text":"P(1, 2)","location":"specification"}]},{"type":"text","order":3,"title":"Find tangent line equation","content":"With the gradient $m = -\\frac{2}{3}$ and the point $P(1, 2)$, we use the point-gradient formula $y - y_1 = m(x - x_1)$:\n\n$$y - 2 = -\\frac{2}{3}(x - 1)$$\n\nMultiplying through by $3$ to remove fractions:\n\n$$3(y - 2) = -2(x - 1)$$\n\n$$3y - 6 = -2x + 2$$\n\nRearranging into standard form gives:\n\n$$2x + 3y = 8$$"}]},{"id":"763d3587-a147-43c9-a9b8-8eed0d9c1ccd:conic_tangent_formula","label":"(b) Conic Tangent Formula","steps":[{"type":"text","order":0,"title":"Introduce the Tangent Rule","content":"For any second-degree conic curve, we can find the equation of the tangent at a point $(x_1, y_1)$ on the curve directly. We do this by replacing the quadratic and linear terms using the standard **$T = 0$ linear substitution rules**:\n\n* $x^2 \\to x_1x$\n* $y^2 \\to y_1y$\n* $xy \\to \\frac{x_1y + y_1x}{2}$\n* $x \\to \\frac{x + x_1}{2}$\n* $y \\to \\frac{y + y_1}{2}$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Curve equation:"},{"latex":"$$3\\htmlData{anchor=x2}{x^2} + \\htmlData{anchor=xy}{xy} - 2\\htmlData{anchor=x}{x} + 2\\htmlData{anchor=y2}{y^2} = 11$"},{"text":"Tangent equation:"},{"latex":"$$3\\htmlData{anchor=x1x}{(x_1x)} + \\htmlData{anchor=xysub}{\\frac{x_1y + y_1x}{2}} - 2\\htmlData{anchor=xsub}{\\left(\\frac{x + x_1}{2}\\right)} + 2\\htmlData{anchor=y1y}{(y_1y)} = 11$"}],"highlights":[{"color":"sky","style":"underline","anchor":"x2"},{"color":"amber","style":"underline","anchor":"xy"},{"color":"purple","style":"underline","anchor":"x"},{"color":"green","style":"underline","anchor":"y2"}],"connections":[{"to":"x1x","from":"x2","color":"sky","label":null,"style":"solid"},{"to":"xysub","from":"xy","color":"amber","label":null,"style":"solid"},{"to":"xsub","from":"x","color":"purple","label":null,"style":"solid"},{"to":"y1y","from":"y2","color":"green","label":null,"style":"solid"}]},"order":1,"title":"Apply substitution rules","description":"We replace each term of our curve equation $3x^2 + xy - 2x + 2y^2 = 11$ with its corresponding tangent substitution."},{"type":"text","order":2,"title":"Substitute the point P(1, 2)","content":"Now, we substitute the coordinates of the point $P(x_1, y_1) = (1, 2)$ into our tangent equation:\n\n$$3(1)x + \\frac{(1)y + (2)x}{2} - 2\\left(\\frac{x + 1}{2}\\right) + 2(2)y = 11$$\n\nSimplifying this term by term:\n\n$$3x + \\frac{y + 2x}{2} - (x + 1) + 4y = 11$$"},{"type":"text","order":3,"title":"Simplify and solve","content":"To eliminate the fraction, multiply the entire equation by $2$:\n\n$$6x + (y + 2x) - 2(x + 1) + 8y = 22$$\n\nExpand and collect like terms:\n\n$$\\begin{aligned} 6x + y + 2x - 2x - 2 + 8y &= 22 \\\\ 6x + 9y - 2 &= 22 \\\\ 6x + 9y &= 24 \\end{aligned}$$\n\nDividing the entire equation by $3$ gives the simplified linear equation:\n\n$$2x + 3y = 8$$"}]}],"generatedAt":"2026-06-09T02:01:47.515Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"5ea19cbc-a34a-43e6-ae77-70f238fbae5e","revisionId":"63001e0d-a982-43af-b9a3-20cb65c0e3d4","questionNumber":null,"questionType":"LA","specification":"The point $K$ lies on the curve with equation $y = \\frac{3e^x+1}{e^x+1}$ with $x$-coordinate 0.","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"1d05a1b1-d6e3-4cc7-8c83-d1676e36b340","label":"a","content":"Find an equation of the normal to the curve at the point $K$.","markscheme":"- Differentiates using the quotient rule: $\\frac{dy}{dx} = \\frac{3e^x(e^x + 1) - e^x(3e^x + 1)}{(e^x + 1)^2} = \\frac{2e^x}{(e^x + 1)^2}$ M1 A1\n- Substitutes $x=0$ to find the gradient of the tangent and then the normal: $\\frac{dy}{dx} = \\frac{2}{4} = \\frac{1}{2} \\Rightarrow m_{\\text{normal}} = -2$ M1\n- Finds the $y$-coordinate at $x=0$: $y = \\frac{3+1}{1+1} = 2$ B1\n- Equation: $y - 2 = -2(x - 0) \\Rightarrow y = -2x + 2$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Selection and application of quotient rule.","code":"M"},{"id":"p2","marks":1,"description":"Correct differentiation result.","code":"A"},{"id":"p3","marks":1,"description":"Calculation of normal gradient via negative reciprocal of tangent gradient.","code":"M"},{"id":"p4","marks":1,"description":"Evaluation of the curve equation at x=0.","code":"B"},{"id":"p5","marks":1,"description":"Final equation of the normal line.","code":"A"}],"finalAnswers":["y = -2x + 2","y + 2x = 2"],"generatedAt":"2026-05-06T14:04:04.721Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"quotient_rule","label":"Quotient Rule Approach","steps":[{"type":"text","order":0,"title":"Identify the goal and method","content":"To find the equation of the normal to the curve at the point $K$ where $x = 0$, we need to:\n\n1. Find the $y$-coordinate of point $K$.\n2. Differentiate the curve's equation using the **Quotient Rule** to find the derivative $\\frac{dy}{dx}$.\n3. Calculate the tangent gradient $m_{\\text{tangent}}$ at $x = 0$.\n4. Determine the perpendicular normal gradient $m_{\\text{normal}} = -\\frac{1}{m_{\\text{tangent}}}$.\n5. Write down the equation of the normal line using the point-gradient formula."},{"type":"text","order":1,"title":"Find the y-coordinate","content":"Let's find the $y$-coordinate of the point $K$ by substituting $x = 0$ into the curve's equation:\n\n$$y = \\frac{3e^0 + 1}{e^0 + 1}$$\n\nSince $e^0 = 1$, we get:\n\n$$y = \\frac{3(1) + 1}{1 + 1} = \\frac{4}{2} = 2$$\n\nSo, the coordinates of the point $K$ are $(0, 2)$."},{"type":"hyper_latex","block":{"nodes":[{"text":"The Quotient Rule states that if a function is a fraction of two parts, its derivative is given by:"},{"latex":"$$\\frac{d}{dx}\\left(\\frac{u}{v}\\right) = \\frac{u'v - uv'}{v^2}$"},{"text":"For our function, we define the numerator and denominator:"},{"latex":"$$\\htmlData{anchor=u-node}{\\color{@sky}{u = 3e^x + 1}} \\implies \\htmlData{anchor=du-node}{\\color{@sky}{u' = 3e^x}}$"},{"latex":"$$\\htmlData{anchor=v-node}{\\color{@purple}{v = e^x + 1}} \\implies \\htmlData{anchor=dv-node}{\\color{@purple}{v' = e^x}}$"},{"text":"Substituting these parts into the Quotient Rule formula gives:"},{"latex":"$$\\frac{dy}{dx} = \\frac{(\\htmlData{anchor=du-sub}{\\color{@sky}{3e^x}})(\\htmlData{anchor=v-sub}{\\color{@purple}{e^x + 1}}) - (\\htmlData{anchor=u-sub}{\\color{@sky}{3e^x + 1}})(\\htmlData{anchor=dv-sub}{\\color{@purple}{e^x}})}{(\\htmlData{anchor=v2-sub}{\\color{@purple}{e^x + 1}})^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"du-node"},{"color":"purple","style":"text-color","anchor":"v-node"}],"connections":[{"to":"du-sub","from":"du-node","color":"sky","label":null,"style":"solid"},{"to":"v-sub","from":"v-node","color":"purple","label":null,"style":"solid"},{"to":"u-sub","from":"u-node","color":"sky","label":null,"style":"solid"},{"to":"dv-sub","from":"dv-node","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Differentiate using quotient rule","description":"We apply the Quotient Rule by setting $u$ as the numerator and $v$ as the denominator."},{"type":"text","order":3,"title":"Simplify the derivative","content":"Let's simplify the numerator of our derivative:\n\n$$3e^x(e^x + 1) - e^x(3e^x + 1) = 3e^{2x} + 3e^x - (3e^{2x} + e^x)$$\n$$= 3e^{2x} + 3e^x - 3e^{2x} - e^x$$\n$$= 2e^x$$\n\nThus, the simplified derivative is:\n\n$$\\frac{dy}{dx} = \\frac{2e^x}{(e^x + 1)^2}$$\n\nThis working helps us earn M1 and A1 for correct differentiation."},{"type":"text","order":4,"title":"Calculate the gradients","content":"First, evaluate the gradient of the tangent at $x = 0$ by substituting $x = 0$ into $\\frac{dy}{dx}$:\n\n$$m_{\\text{tangent}} = \\frac{2e^0}{(e^0 + 1)^2} = \\frac{2(1)}{(1 + 1)^2} = \\frac{2}{4} = \\frac{1}{2}$$\n\nSince the normal is perpendicular to the tangent, its gradient is the negative reciprocal:\n\n$$m_{\\text{normal}} = -\\frac{1}{m_{\\text{tangent}}} = -2$$\n\nThis step earns the third mark M1."},{"type":"text","order":5,"title":"Find the normal's equation","content":"We now have the gradient of the normal $m = -2$ and the point $K(0, 2)$.\n\nUsing the equation of a straight line, $y - y_1 = m(x - x_1)$:\n\n$$y - 2 = -2(x - 0)$$\n\n$$y - 2 = -2x$$\n\n$$y = -2x + 2$$\n\nThis leads to the final equation, earning the final mark A1."}]},{"id":"algebraic_simplification","label":"Algebraic Simplification and Chain Rule","steps":[{"type":"text","order":0,"title":"Find the coordinate at K","content":"To find the coordinates of point $K$, we substitute $x = 0$ into the equation of the curve:\n\n$$y = \\frac{3e^x + 1}{e^x + 1}$$\n\nSubstituting $x = 0$ (recalling that $e^0 = 1$):\n\n$$y = \\frac{3(1) + 1}{1 + 1} = \\frac{4}{2} = 2$$\n\nSo, the coordinates of the point $K$ are $(0, 2)$. This earns you the first mark B1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{3e^x + 1}{e^x + 1}$"},{"latex":"$$y = \\frac{\\htmlData{anchor=num-split}{\\color{@sky}{3(e^x + 1) - 2}}}{e^x + 1}$"},{"latex":"$$y = \\htmlData{anchor=frac-split}{\\color{@amber}{3 - 2(e^x + 1)^{-1}}}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"num-split"},{"color":"amber","style":"box","anchor":"frac-split"}],"connections":[{"to":"frac-split","from":"num-split","color":"amber","label":"divide each term","style":"solid"}]},"order":1,"title":"Simplify the equation algebraically","description":"Before differentiating, we can rewrite the function algebraically so that we do not have to use the quotient rule. We express the numerator in terms of the denominator: $3e^x + 1 = 3(e^x + 1) - 2$."},{"type":"text","order":2,"title":"Differentiate using Chain Rule","content":"Now, we differentiate $y = 3 - 2(e^x + 1)^{-1}$ with respect to $x$ using the chain rule. \n\nLet the outer function be $u^{-1}$ and the inner function be $u = e^x + 1$. The derivative is:\n\n$$\\frac{dy}{dx} = 0 - 2 \\cdot (-1)(e^x + 1)^{-2} \\cdot \\frac{d}{dx}(e^x + 1)$$\n\n$$\\frac{dy}{dx} = 2(e^x + 1)^{-2} \\cdot e^x = \\frac{2e^x}{(e^x + 1)^2}$$\n\nThis simplified derivative matches the one obtained from the quotient rule, earning you M1 A1."},{"type":"text","order":3,"title":"Find the normal gradient","content":"First, find the gradient of the tangent at $x = 0$ by evaluating the derivative:\n\n$$m_{\\text{tangent}} = \\frac{2e^0}{(e^0 + 1)^2} = \\frac{2(1)}{(1 + 1)^2} = \\frac{2}{4} = \\frac{1}{2}$$\n\nSince the normal is perpendicular to the tangent, their gradients satisfy $m_{\\text{normal}} \\times m_{\\text{tangent}} = -1$:\n\n$$m_{\\text{normal}} = -\\frac{1}{m_{\\text{tangent}}} = -2$$\n\nThis step earns you M1."},{"type":"text","order":4,"title":"Determine equation of normal","content":"Using the equation of a straight line $y - y_1 = m(x - x_1)$ with our point $K(0, 2)$ and $m_{\\text{normal}} = -2$:\n\n$$y - 2 = -2(x - 0)$$\n\n$$y = -2x + 2$$\n\nThis gives us our final answer, earning the final mark A1."}]},{"id":"product_rule","label":"Product Rule Approach","steps":[{"type":"text","order":0,"title":"Find the $y$-coordinate of $K$","content":"To find the coordinates of point $K$, we substitute the $x$-coordinate, which is $x = 0$, into the equation of the curve:\n\n$$y = \\frac{3e^x + 1}{e^x + 1}$$\n\nSubstituting $x = 0$:\n\n$$y = \\frac{3e^0 + 1}{e^0 + 1} = \\frac{3(1) + 1}{1 + 1} = \\frac{4}{2} = 2$$\n\nSo, point $K$ has coordinates $(0, 2)$. This earns us our first mark."},{"type":"text","order":1,"title":"Rewrite for Product Rule","content":"Using our chosen method, we express the fraction as a product of two functions so we can apply the Product Rule:\n\n$$y = (3e^x + 1)(e^x + 1)^{-1}$$\n\nLet's define:\n- $u = 3e^x + 1 \\implies \\frac{du}{dx} = 3e^x$\n- $v = (e^x + 1)^{-1}$"},{"type":"hyper_latex","block":{"nodes":[{"text":"To find the derivative of the reciprocal term, we apply the chain rule:"},{"latex":"$$\\htmlData{anchor=v-func}{v} = (\\htmlData{anchor=v-inner}{\\color{@sky}{e^x + 1}})^{-1}$"},{"latex":"$$\\htmlData{anchor=v-deriv}{\\frac{dv}{dx}} = -(\\color{@sky}{e^x + 1})^{-2} \\times \\htmlData{anchor=v-inner-deriv}{\\color{@amber}{e^x}}$"},{"latex":"$$\\phantom{\\frac{dv}{dx}} = -\\frac{\\color{@amber}{e^x}}{(\\color{@sky}{e^x + 1})^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"v-inner"},{"color":"amber","style":"text-color","anchor":"v-inner-deriv"}],"connections":[{"to":"v-inner-deriv","from":"v-inner","color":"amber","label":"derivative of inner function","style":"solid"}]},"order":2,"title":"Differentiate v using Chain Rule","description":"We apply the chain rule: differentiate the outer power function first, then multiply by the derivative of the inner function $e^x + 1$."},{"type":"text","order":3,"title":"Apply the Product Rule","content":"Now we use the product rule, which is given by:\n\n$$\\frac{dy}{dx} = u\\frac{dv}{dx} + v\\frac{du}{dx}$$\n\nSubstituting our terms:\n\n$$\\frac{dy}{dx} = (3e^x + 1)\\left(-\\frac{e^x}{(e^x + 1)^2}\\right) + (e^x + 1)^{-1}(3e^x)$$\n\n$$\\frac{dy}{dx} = -\\frac{e^x(3e^x + 1)}{(e^x + 1)^2} + \\frac{3e^x}{e^x + 1}$$\n\nTo combine them, find a common denominator of $(e^x + 1)^2$:\n\n$$\\frac{dy}{dx} = \\frac{-e^x(3e^x + 1) + 3e^x(e^x + 1)}{(e^x + 1)^2}$$\n\n$$\\frac{dy}{dx} = \\frac{-3e^{2x} - e^x + 3e^{2x} + 3e^x}{(e^x + 1)^2} = \\frac{2e^x}{(e^x + 1)^2}$$\n\nThis confirms the derivative obtained using our product rule approach, earning M1 and A1."},{"type":"text","order":4,"title":"Calculate the normal gradient","content":"Next, evaluate the derivative at $x = 0$ to find the gradient of the tangent, $m_{\\text{tangent}}$:\n\n$$m_{\\text{tangent}} = \\frac{2e^0}{(e^0 + 1)^2} = \\frac{2(1)}{(1 + 1)^2} = \\frac{2}{4} = \\frac{1}{2}$$\n\nSince the normal is perpendicular to the tangent, its gradient $m_{\\text{normal}}$ is the negative reciprocal of $m_{\\text{tangent}}$:\n\n$$m_{\\text{normal}} = -\\frac{1}{m_{\\text{tangent}}} = -2$$\n\nThis step earns M1."},{"type":"text","order":5,"title":"Find the normal equation","content":"Finally, we find the equation of the normal line at point $K(0, 2)$ with gradient $m = -2$ using the line equation formula:\n\n$$y - y_1 = m(x - x_1)$$\n\n$$y - 2 = -2(x - 0)$$\n\n$$y = -2x + 2$$\n\nThis completes our proof and gives us the final equation, earning A1."}]}],"generatedAt":"2026-06-09T05:35:58.945Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"1d05a1b1-d6e3-4cc7-8c83-d1676e36b340:quotient_rule","label":"Quotient Rule Approach","steps":[{"type":"text","order":0,"title":"Identify the goal and method","content":"To find the equation of the normal to the curve at the point $K$ where $x = 0$, we need to:\n\n1. Find the $y$-coordinate of point $K$.\n2. Differentiate the curve's equation using the **Quotient Rule** to find the derivative $\\frac{dy}{dx}$.\n3. Calculate the tangent gradient $m_{\\text{tangent}}$ at $x = 0$.\n4. Determine the perpendicular normal gradient $m_{\\text{normal}} = -\\frac{1}{m_{\\text{tangent}}}$.\n5. Write down the equation of the normal line using the point-gradient formula."},{"type":"text","order":1,"title":"Find the y-coordinate","content":"Let's find the $y$-coordinate of the point $K$ by substituting $x = 0$ into the curve's equation:\n\n$$y = \\frac{3e^0 + 1}{e^0 + 1}$$\n\nSince $e^0 = 1$, we get:\n\n$$y = \\frac{3(1) + 1}{1 + 1} = \\frac{4}{2} = 2$$\n\nSo, the coordinates of the point $K$ are $(0, 2)$."},{"type":"hyper_latex","block":{"nodes":[{"text":"The Quotient Rule states that if a function is a fraction of two parts, its derivative is given by:"},{"latex":"$$\\frac{d}{dx}\\left(\\frac{u}{v}\\right) = \\frac{u'v - uv'}{v^2}$"},{"text":"For our function, we define the numerator and denominator:"},{"latex":"$$\\htmlData{anchor=u-node}{\\color{@sky}{u = 3e^x + 1}} \\implies \\htmlData{anchor=du-node}{\\color{@sky}{u' = 3e^x}}$"},{"latex":"$$\\htmlData{anchor=v-node}{\\color{@purple}{v = e^x + 1}} \\implies \\htmlData{anchor=dv-node}{\\color{@purple}{v' = e^x}}$"},{"text":"Substituting these parts into the Quotient Rule formula gives:"},{"latex":"$$\\frac{dy}{dx} = \\frac{(\\htmlData{anchor=du-sub}{\\color{@sky}{3e^x}})(\\htmlData{anchor=v-sub}{\\color{@purple}{e^x + 1}}) - (\\htmlData{anchor=u-sub}{\\color{@sky}{3e^x + 1}})(\\htmlData{anchor=dv-sub}{\\color{@purple}{e^x}})}{(\\htmlData{anchor=v2-sub}{\\color{@purple}{e^x + 1}})^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"du-node"},{"color":"purple","style":"text-color","anchor":"v-node"}],"connections":[{"to":"du-sub","from":"du-node","color":"sky","label":null,"style":"solid"},{"to":"v-sub","from":"v-node","color":"purple","label":null,"style":"solid"},{"to":"u-sub","from":"u-node","color":"sky","label":null,"style":"solid"},{"to":"dv-sub","from":"dv-node","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Differentiate using quotient rule","description":"We apply the Quotient Rule by setting $u$ as the numerator and $v$ as the denominator."},{"type":"text","order":3,"title":"Simplify the derivative","content":"Let's simplify the numerator of our derivative:\n\n$$3e^x(e^x + 1) - e^x(3e^x + 1) = 3e^{2x} + 3e^x - (3e^{2x} + e^x)$$\n$$= 3e^{2x} + 3e^x - 3e^{2x} - e^x$$\n$$= 2e^x$$\n\nThus, the simplified derivative is:\n\n$$\\frac{dy}{dx} = \\frac{2e^x}{(e^x + 1)^2}$$\n\nThis working helps us earn M1 and A1 for correct differentiation."},{"type":"text","order":4,"title":"Calculate the gradients","content":"First, evaluate the gradient of the tangent at $x = 0$ by substituting $x = 0$ into $\\frac{dy}{dx}$:\n\n$$m_{\\text{tangent}} = \\frac{2e^0}{(e^0 + 1)^2} = \\frac{2(1)}{(1 + 1)^2} = \\frac{2}{4} = \\frac{1}{2}$$\n\nSince the normal is perpendicular to the tangent, its gradient is the negative reciprocal:\n\n$$m_{\\text{normal}} = -\\frac{1}{m_{\\text{tangent}}} = -2$$\n\nThis step earns the third mark M1."},{"type":"text","order":5,"title":"Find the normal's equation","content":"We now have the gradient of the normal $m = -2$ and the point $K(0, 2)$.\n\nUsing the equation of a straight line, $y - y_1 = m(x - x_1)$:\n\n$$y - 2 = -2(x - 0)$$\n\n$$y - 2 = -2x$$\n\n$$y = -2x + 2$$\n\nThis leads to the final equation, earning the final mark A1."}]},{"id":"1d05a1b1-d6e3-4cc7-8c83-d1676e36b340:algebraic_simplification","label":"Algebraic Simplification and Chain Rule","steps":[{"type":"text","order":0,"title":"Find the coordinate at K","content":"To find the coordinates of point $K$, we substitute $x = 0$ into the equation of the curve:\n\n$$y = \\frac{3e^x + 1}{e^x + 1}$$\n\nSubstituting $x = 0$ (recalling that $e^0 = 1$):\n\n$$y = \\frac{3(1) + 1}{1 + 1} = \\frac{4}{2} = 2$$\n\nSo, the coordinates of the point $K$ are $(0, 2)$. This earns you the first mark B1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{3e^x + 1}{e^x + 1}$"},{"latex":"$$y = \\frac{\\htmlData{anchor=num-split}{\\color{@sky}{3(e^x + 1) - 2}}}{e^x + 1}$"},{"latex":"$$y = \\htmlData{anchor=frac-split}{\\color{@amber}{3 - 2(e^x + 1)^{-1}}}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"num-split"},{"color":"amber","style":"box","anchor":"frac-split"}],"connections":[{"to":"frac-split","from":"num-split","color":"amber","label":"divide each term","style":"solid"}]},"order":1,"title":"Simplify the equation algebraically","description":"Before differentiating, we can rewrite the function algebraically so that we do not have to use the quotient rule. We express the numerator in terms of the denominator: $3e^x + 1 = 3(e^x + 1) - 2$."},{"type":"text","order":2,"title":"Differentiate using Chain Rule","content":"Now, we differentiate $y = 3 - 2(e^x + 1)^{-1}$ with respect to $x$ using the chain rule. \n\nLet the outer function be $u^{-1}$ and the inner function be $u = e^x + 1$. The derivative is:\n\n$$\\frac{dy}{dx} = 0 - 2 \\cdot (-1)(e^x + 1)^{-2} \\cdot \\frac{d}{dx}(e^x + 1)$$\n\n$$\\frac{dy}{dx} = 2(e^x + 1)^{-2} \\cdot e^x = \\frac{2e^x}{(e^x + 1)^2}$$\n\nThis simplified derivative matches the one obtained from the quotient rule, earning you M1 A1."},{"type":"text","order":3,"title":"Find the normal gradient","content":"First, find the gradient of the tangent at $x = 0$ by evaluating the derivative:\n\n$$m_{\\text{tangent}} = \\frac{2e^0}{(e^0 + 1)^2} = \\frac{2(1)}{(1 + 1)^2} = \\frac{2}{4} = \\frac{1}{2}$$\n\nSince the normal is perpendicular to the tangent, their gradients satisfy $m_{\\text{normal}} \\times m_{\\text{tangent}} = -1$:\n\n$$m_{\\text{normal}} = -\\frac{1}{m_{\\text{tangent}}} = -2$$\n\nThis step earns you M1."},{"type":"text","order":4,"title":"Determine equation of normal","content":"Using the equation of a straight line $y - y_1 = m(x - x_1)$ with our point $K(0, 2)$ and $m_{\\text{normal}} = -2$:\n\n$$y - 2 = -2(x - 0)$$\n\n$$y = -2x + 2$$\n\nThis gives us our final answer, earning the final mark A1."}]},{"id":"1d05a1b1-d6e3-4cc7-8c83-d1676e36b340:product_rule","label":"Product Rule Approach","steps":[{"type":"text","order":0,"title":"Find the $y$-coordinate of $K$","content":"To find the coordinates of point $K$, we substitute the $x$-coordinate, which is $x = 0$, into the equation of the curve:\n\n$$y = \\frac{3e^x + 1}{e^x + 1}$$\n\nSubstituting $x = 0$:\n\n$$y = \\frac{3e^0 + 1}{e^0 + 1} = \\frac{3(1) + 1}{1 + 1} = \\frac{4}{2} = 2$$\n\nSo, point $K$ has coordinates $(0, 2)$. This earns us our first mark."},{"type":"text","order":1,"title":"Rewrite for Product Rule","content":"Using our chosen method, we express the fraction as a product of two functions so we can apply the Product Rule:\n\n$$y = (3e^x + 1)(e^x + 1)^{-1}$$\n\nLet's define:\n- $u = 3e^x + 1 \\implies \\frac{du}{dx} = 3e^x$\n- $v = (e^x + 1)^{-1}$"},{"type":"hyper_latex","block":{"nodes":[{"text":"To find the derivative of the reciprocal term, we apply the chain rule:"},{"latex":"$$\\htmlData{anchor=v-func}{v} = (\\htmlData{anchor=v-inner}{\\color{@sky}{e^x + 1}})^{-1}$"},{"latex":"$$\\htmlData{anchor=v-deriv}{\\frac{dv}{dx}} = -(\\color{@sky}{e^x + 1})^{-2} \\times \\htmlData{anchor=v-inner-deriv}{\\color{@amber}{e^x}}$"},{"latex":"$$\\phantom{\\frac{dv}{dx}} = -\\frac{\\color{@amber}{e^x}}{(\\color{@sky}{e^x + 1})^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"v-inner"},{"color":"amber","style":"text-color","anchor":"v-inner-deriv"}],"connections":[{"to":"v-inner-deriv","from":"v-inner","color":"amber","label":"derivative of inner function","style":"solid"}]},"order":2,"title":"Differentiate v using Chain Rule","description":"We apply the chain rule: differentiate the outer power function first, then multiply by the derivative of the inner function $e^x + 1$."},{"type":"text","order":3,"title":"Apply the Product Rule","content":"Now we use the product rule, which is given by:\n\n$$\\frac{dy}{dx} = u\\frac{dv}{dx} + v\\frac{du}{dx}$$\n\nSubstituting our terms:\n\n$$\\frac{dy}{dx} = (3e^x + 1)\\left(-\\frac{e^x}{(e^x + 1)^2}\\right) + (e^x + 1)^{-1}(3e^x)$$\n\n$$\\frac{dy}{dx} = -\\frac{e^x(3e^x + 1)}{(e^x + 1)^2} + \\frac{3e^x}{e^x + 1}$$\n\nTo combine them, find a common denominator of $(e^x + 1)^2$:\n\n$$\\frac{dy}{dx} = \\frac{-e^x(3e^x + 1) + 3e^x(e^x + 1)}{(e^x + 1)^2}$$\n\n$$\\frac{dy}{dx} = \\frac{-3e^{2x} - e^x + 3e^{2x} + 3e^x}{(e^x + 1)^2} = \\frac{2e^x}{(e^x + 1)^2}$$\n\nThis confirms the derivative obtained using our product rule approach, earning M1 and A1."},{"type":"text","order":4,"title":"Calculate the normal gradient","content":"Next, evaluate the derivative at $x = 0$ to find the gradient of the tangent, $m_{\\text{tangent}}$:\n\n$$m_{\\text{tangent}} = \\frac{2e^0}{(e^0 + 1)^2} = \\frac{2(1)}{(1 + 1)^2} = \\frac{2}{4} = \\frac{1}{2}$$\n\nSince the normal is perpendicular to the tangent, its gradient $m_{\\text{normal}}$ is the negative reciprocal of $m_{\\text{tangent}}$:\n\n$$m_{\\text{normal}} = -\\frac{1}{m_{\\text{tangent}}} = -2$$\n\nThis step earns M1."},{"type":"text","order":5,"title":"Find the normal equation","content":"Finally, we find the equation of the normal line at point $K(0, 2)$ with gradient $m = -2$ using the line equation formula:\n\n$$y - y_1 = m(x - x_1)$$\n\n$$y - 2 = -2(x - 0)$$\n\n$$y = -2x + 2$$\n\nThis completes our proof and gives us the final equation, earning A1."}]}],"generatedAt":"2026-06-09T05:36:11.845Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"60601d3d-8f39-40ad-a875-04eae3ad5f41","revisionId":"19f1ff97-6668-4c8f-8ee4-547b87617183","questionNumber":null,"questionType":"LA","specification":"Differentiate with respect to $x$,","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"b1fe77e3-2ae4-4b96-9404-9643f868353c","label":"a","content":"$$x^2 \\sqrt{3x+5}$","markscheme":"- Differentiates $\\sqrt{3x+5}$ to get $\\frac{3}{2}(3x+5)^{-\\frac{1}{2}}$ using the chain rule M1\n- Applies product rule: $2x(3x+5)^{\\frac{1}{2}} + \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}}$ M1\n- Correct derivative expression A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct chain rule for root term.","code":"M"},{"id":"p2","marks":1,"description":"Correct setup of product rule.","code":"M"},{"id":"p3","marks":1,"description":"Correct simplified or unsimplified derivative.","code":"A"}],"finalAnswers":["$$2x(3x+5)^{\\frac{1}{2}} + \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}}$"],"generatedAt":"2026-05-06T14:04:01.964Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"product_chain_rule","label":"Product and Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the rules needed","content":"We need to differentiate the function $$y = x^2 \\sqrt{3x+5}$$ with respect to $x$.\n\nSince this is a product of two terms, $u(x) = x^2$ and $v(x) = \\sqrt{3x+5}$, we will use the **Product Rule**:\n$$\\frac{d}{dx}(uv) = u'v + uv'$$\n\nTo differentiate $v(x) = (3x+5)^{\\frac{1}{2}}$, we will also need to use the **Chain Rule**."},{"type":"text","order":1,"title":"Differentiate the individual parts","content":"Let's find the derivatives of $u$ and $v$ separately:\n\n1. For $u = x^2$:\n$$u' = \\frac{du}{dx} = 2x$$\n\n2. For $v = (3x+5)^{\\frac{1}{2}}$, we apply the chain rule:\n$$\\begin{aligned} v' = \\frac{dv}{dx} &= \\frac{1}{2}(3x+5)^{-\\frac{1}{2}} \\cdot \\frac{d}{dx}(3x+5) \\\\ &= \\frac{1}{2}(3x+5)^{-\\frac{1}{2}} \\cdot 3 \\\\ &= \\frac{3}{2}(3x+5)^{-\\frac{1}{2}} \\end{aligned}$$\n\nThis earns the first mark M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-term}{x^2}, \\quad v = \\htmlData{anchor=v-term}{\\sqrt{3x+5}}$"},{"latex":"$$u' = \\htmlData{anchor=u-deriv}{2x}, \\quad v' = \\htmlData{anchor=v-deriv}{\\frac{3}{2}(3x+5)^{-\\frac{1}{2}}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=u-deriv-sub}{\\color{@sky}{2x}}\\htmlData{anchor=v-term-sub}{\\color{@pink}{(3x+5)^{\\frac{1}{2}}}} + \\htmlData{anchor=u-term-sub}{\\color{@amber}{x^2}}\\left(\\htmlData{anchor=v-deriv-sub}{\\color{@teal}{\\frac{3}{2}(3x+5)^{-\\frac{1}{2}}}}\\right)$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-deriv-sub"},{"color":"pink","style":"text-color","anchor":"v-term-sub"},{"color":"amber","style":"text-color","anchor":"u-term-sub"},{"color":"teal","style":"text-color","anchor":"v-deriv-sub"}],"connections":[{"to":"u-deriv-sub","from":"u-deriv","color":"sky","label":"u'","style":"solid"},{"to":"v-term-sub","from":"v-term","color":"pink","label":"v","style":"solid"},{"to":"u-term-sub","from":"u-term","color":"amber","label":"u","style":"solid"},{"to":"v-deriv-sub","from":"v-deriv","color":"teal","label":"v'","style":"solid"}]},"order":2,"title":"Apply the Product Rule","description":"Now, we substitute $u$, $u'$, $v$, and $v'$ into the Product Rule formula: $u'v + uv'$. This step earns the second mark M1."},{"type":"text","order":3,"title":"Simplify the final expression","content":"Combining the terms, we get our final derivative expression:\n\n$$\\frac{dy}{dx} = 2x(3x+5)^{\\frac{1}{2}} + \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}}$$\n\nThis fully correct final expression completes our work and earns the final mark A1."}]},{"id":"single_radical","label":"Rewriting Under a Single Radical","steps":[{"type":"text","order":0,"title":"Rewrite under one radical","content":"To rewrite the expression $y = x^2 \\sqrt{3x+5}$ under a single radical, we first convert the outside term, $x^2$, into a square root form:\n\n$$x^2 = \\sqrt{x^4}$$\n\nNow, we can combine this with the existing square root term:\n\n$$\\begin{aligned} y &= \\sqrt{x^4} \\cdot \\sqrt{3x+5} \\\\ &= \\sqrt{x^4(3x+5)} \\\\ &= \\sqrt{3x^5 + 5x^4} \\end{aligned}$$"},{"type":"text","order":1,"title":"Express with fractional power","content":"To prepare the expression for differentiation, we rewrite the square root as a fractional exponent:\n\n$$y = (3x^5 + 5x^4)^{\\frac{1}{2}}$$"},{"type":"text","order":2,"title":"Apply the chain rule","content":"We now differentiate using the **chain rule**.\n\nLet the inner function be $u = 3x^5 + 5x^4$, so $y = u^{\\frac{1}{2}}$.\n\nThe chain rule states:\n\n$$\\frac{dy}{dx} = \\frac{dy}{du} \\cdot \\frac{du}{dx}$$\n\n1. Find $\\frac{dy}{du}$:\n$$\\frac{dy}{du} = \\frac{1}{2}u^{-\\frac{1}{2}}$$\n\n2. Find $\\frac{du}{dx}$:\n$$\\frac{du}{dx} = 15x^4 + 20x^3$$"},{"type":"text","order":3,"title":"Combine and simplify","content":"Substitute these parts back into the chain rule formula:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= \\frac{1}{2}(3x^5 + 5x^4)^{-\\frac{1}{2}} \\cdot (15x^4 + 20x^3) \\\\ &= \\frac{15x^4 + 20x^3}{2\\sqrt{3x^5 + 5x^4}} \\end{aligned}$$\n\nWe can factor out $x^4$ from the radical in the denominator to simplify the final fraction:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= \\frac{x^3(15x + 20)}{2\\sqrt{x^4(3x + 5)}} \\\\ &= \\frac{x^3(15x + 20)}{2x^2\\sqrt{3x+5}} \\\\ &= \\frac{x(15x+20)}{2\\sqrt{3x+5}} \\\\ &= \\frac{15x^2 + 20x}{2\\sqrt{3x+5}} \\end{aligned}$$"},{"type":"text","order":4,"title":"Verify equivalence","content":"To confirm this matches the standard product rule form from the markscheme, we can split our answer:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= \\frac{12x^2 + 20x + 3x^2}{2\\sqrt{3x+5}} \\\\ &= \\frac{4x(3x+5)}{2\\sqrt{3x+5}} + \\frac{3x^2}{2\\sqrt{3x+5}} \\\\ &= 2x\\sqrt{3x+5} + \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}} \\end{aligned}$$\n\nBoth forms are completely correct and earn full marks!"}]},{"id":"logarithmic_differentiation","label":"Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Define the function","content":"To differentiate the function using **logarithmic differentiation**, we first set the function equal to $y$:\n\n$$y = x^2 \\sqrt{3x+5}$$\n\nThis method is particularly elegant for handling products and roots, as taking the natural logarithm allows us to break the expression down into simpler, individual terms."},{"type":"text","order":1,"title":"Take the logarithm","content":"Take the natural logarithm ($\\ln$) of both sides of the equation:\n\n$$\\ln(y) = \\ln\\left(x^2 \\sqrt{3x+5}\\right)$$"},{"type":"text","order":2,"title":"Simplify using log laws","content":"Using the properties of logarithms, specifically:\n- Product rule: $\\ln(ab) = \\ln(a) + \\ln(b)$\n- Power rule: $\\ln(a^n) = n \\ln(a)$\n\nWe can rewrite the right-hand side. First, write the square root as a fractional power:\n\n$$\\ln(y) = \\ln\\left(x^2 (3x+5)^{\\frac{1}{2}}\\right)$$\n\nNow, expand the logarithm:\n\n$$\\ln(y) = \\ln(x^2) + \\ln\\left((3x+5)^{\\frac{1}{2}}\\right)$$\n\n$$\\ln(y) = 2\\ln(x) + \\frac{1}{2}\\ln(3x+5)$$"},{"type":"text","order":3,"title":"Differentiate implicitly","content":"Now we differentiate both sides with respect to $x$ using the chain rule.\n\nOn the left-hand side, using implicit differentiation:\n\n$$\\frac{d}{dx}[\\ln(y)] = \\frac{1}{y} \\frac{dy}{dx}$$\n\nOn the right-hand side, we differentiate each term:\n- The derivative of $2\\ln(x)$ is $\\frac{2}{x}$.\n- For $\\frac{1}{2}\\ln(3x+5)$, we use the chain rule: $\\frac{1}{2} \\cdot \\frac{1}{3x+5} \\cdot 3 = \\frac{3}{2(3x+5)}$.\n\nPutting this together, we get:\n\n$$\\frac{1}{y} \\frac{dy}{dx} = \\frac{2}{x} + \\frac{3}{2(3x+5)}$$"},{"type":"text","order":4,"title":"Solve for dy/dx","content":"Multiply both sides by $y$ to isolate the derivative $\\frac{dy}{dx}$:\n\n$$\\frac{dy}{dx} = y \\left[ \\frac{2}{x} + \\frac{3}{2(3x+5)} \\right]$$\n\nNow substitute the original expression for $y = x^2\\sqrt{3x+5}$ back into the equation:\n\n$$\\frac{dy}{dx} = x^2\\sqrt{3x+5} \\left[ \\frac{2}{x} + \\frac{3}{2(3x+5)} \\right]$$"},{"type":"text","order":5,"title":"Simplify the expression","content":"To get our final derivative expression, we distribute the $x^2\\sqrt{3x+5}$ terms:\n\n$$\\frac{dy}{dx} = \\frac{2x^2\\sqrt{3x+5}}{x} + \\frac{3x^2\\sqrt{3x+5}}{2(3x+5)}$$\n\nSimplify each term:\n- For the first term: $\\frac{2x^2\\sqrt{3x+5}}{x} = 2x\\sqrt{3x+5} = 2x(3x+5)^{\\frac{1}{2}}$\n- For the second term, write the denominator as $(3x+5)^1$ and use the laws of indices: $\\frac{(3x+5)^{\\frac{1}{2}}}{(3x+5)^1} = (3x+5)^{-\\frac{1}{2}}$:\n\n$$\\frac{3x^2\\sqrt{3x+5}}{2(3x+5)} = \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}}$$\n\nThus, we get our final differentiated expression:\n\n$$\\frac{dy}{dx} = 2x(3x+5)^{\\frac{1}{2}} + \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}}$$"}]}],"generatedAt":"2026-06-09T06:10:03.033Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"58d96fce-5e03-4c41-b54e-2b57b7f96934","label":"b","content":"$$x \\ln(4x-1)$","markscheme":"- Differentiates $\\ln(4x-1)$ to get $\\frac{4}{4x-1}$ M1\n- Applies product rule: $1 \\cdot \\ln(4x-1) + x \\cdot \\frac{4}{4x-1}$ M1\n- Correct expression $\\ln(4x-1) + \\frac{4x}{4x-1}$ A1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Differentiates $\\ln(ax+b)$ correctly.","code":"M"},{"id":"p2","marks":1,"description":"Correct application of product rule.","code":"M"},{"id":"p3","marks":1,"description":"Correct final derivative.","code":"A"}],"finalAnswers":["$$\\ln(4x-1) + \\frac{4x}{4x-1}$"],"generatedAt":"2026-05-06T14:04:01.964Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"product_chain_rule","label":"Product and Chain Rules","steps":[{"type":"text","order":0,"title":"Identify the product rule","content":"We need to differentiate the function $y = x \\ln(4x-1)$. Since this function is the product of two simpler functions, we will use the **Product Rule**: $$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$ Let's define our two parts: $u = x$ and $v = \\ln(4x-1)$."},{"type":"text","order":1,"title":"Differentiate each part","content":"Now we differentiate each part with respect to $x$. For $u = x$, we have $$\\frac{du}{dx} = 1$$ For $v = \\ln(4x-1)$, we use the **Chain Rule**: $$\\frac{dv}{dx} = \\frac{1}{4x-1} \\cdot \\frac{d}{dx}(4x-1) = \\frac{4}{4x-1}$$ This earns the first method mark **M1**."},{"type":"text","order":2,"title":"Apply product rule formula","content":"Substitute our components into the Product Rule formula: $$\\frac{dy}{dx} = u\\frac{dv}{dx} + v\\frac{du}{dx}$$ This gives: $$\\frac{dy}{dx} = x \\cdot \\left(\\frac{4}{4x-1}\\right) + \\ln(4x-1) \\cdot 1$$ This earns the second method mark **M1**."},{"type":"text","order":3,"title":"Simplify the final expression","content":"Simplify the expression to write it in its cleanest form: $$\\frac{dy}{dx} = \\ln(4x-1) + \\frac{4x}{4x-1}$$ This earns the final accuracy mark **A1**."}]}],"generatedAt":"2026-06-09T06:11:23.978Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"b1fe77e3-2ae4-4b96-9404-9643f868353c:product_chain_rule","label":"(a) Product and Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the rules needed","content":"We need to differentiate the function $$y = x^2 \\sqrt{3x+5}$$ with respect to $x$.\n\nSince this is a product of two terms, $u(x) = x^2$ and $v(x) = \\sqrt{3x+5}$, we will use the **Product Rule**:\n$$\\frac{d}{dx}(uv) = u'v + uv'$$\n\nTo differentiate $v(x) = (3x+5)^{\\frac{1}{2}}$, we will also need to use the **Chain Rule**."},{"type":"text","order":1,"title":"Differentiate the individual parts","content":"Let's find the derivatives of $u$ and $v$ separately:\n\n1. For $u = x^2$:\n$$u' = \\frac{du}{dx} = 2x$$\n\n2. For $v = (3x+5)^{\\frac{1}{2}}$, we apply the chain rule:\n$$\\begin{aligned} v' = \\frac{dv}{dx} &= \\frac{1}{2}(3x+5)^{-\\frac{1}{2}} \\cdot \\frac{d}{dx}(3x+5) \\\\ &= \\frac{1}{2}(3x+5)^{-\\frac{1}{2}} \\cdot 3 \\\\ &= \\frac{3}{2}(3x+5)^{-\\frac{1}{2}} \\end{aligned}$$\n\nThis earns the first mark M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-term}{x^2}, \\quad v = \\htmlData{anchor=v-term}{\\sqrt{3x+5}}$"},{"latex":"$$u' = \\htmlData{anchor=u-deriv}{2x}, \\quad v' = \\htmlData{anchor=v-deriv}{\\frac{3}{2}(3x+5)^{-\\frac{1}{2}}}$"},{"latex":"$$\\frac{dy}{dx} = \\htmlData{anchor=u-deriv-sub}{\\color{@sky}{2x}}\\htmlData{anchor=v-term-sub}{\\color{@pink}{(3x+5)^{\\frac{1}{2}}}} + \\htmlData{anchor=u-term-sub}{\\color{@amber}{x^2}}\\left(\\htmlData{anchor=v-deriv-sub}{\\color{@teal}{\\frac{3}{2}(3x+5)^{-\\frac{1}{2}}}}\\right)$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-deriv-sub"},{"color":"pink","style":"text-color","anchor":"v-term-sub"},{"color":"amber","style":"text-color","anchor":"u-term-sub"},{"color":"teal","style":"text-color","anchor":"v-deriv-sub"}],"connections":[{"to":"u-deriv-sub","from":"u-deriv","color":"sky","label":"u'","style":"solid"},{"to":"v-term-sub","from":"v-term","color":"pink","label":"v","style":"solid"},{"to":"u-term-sub","from":"u-term","color":"amber","label":"u","style":"solid"},{"to":"v-deriv-sub","from":"v-deriv","color":"teal","label":"v'","style":"solid"}]},"order":2,"title":"Apply the Product Rule","description":"Now, we substitute $u$, $u'$, $v$, and $v'$ into the Product Rule formula: $u'v + uv'$. This step earns the second mark M1."},{"type":"text","order":3,"title":"Simplify the final expression","content":"Combining the terms, we get our final derivative expression:\n\n$$\\frac{dy}{dx} = 2x(3x+5)^{\\frac{1}{2}} + \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}}$$\n\nThis fully correct final expression completes our work and earns the final mark A1."}]},{"id":"b1fe77e3-2ae4-4b96-9404-9643f868353c:single_radical","label":"(a) Rewriting Under a Single Radical","steps":[{"type":"text","order":0,"title":"Rewrite under one radical","content":"To rewrite the expression $y = x^2 \\sqrt{3x+5}$ under a single radical, we first convert the outside term, $x^2$, into a square root form:\n\n$$x^2 = \\sqrt{x^4}$$\n\nNow, we can combine this with the existing square root term:\n\n$$\\begin{aligned} y &= \\sqrt{x^4} \\cdot \\sqrt{3x+5} \\\\ &= \\sqrt{x^4(3x+5)} \\\\ &= \\sqrt{3x^5 + 5x^4} \\end{aligned}$$"},{"type":"text","order":1,"title":"Express with fractional power","content":"To prepare the expression for differentiation, we rewrite the square root as a fractional exponent:\n\n$$y = (3x^5 + 5x^4)^{\\frac{1}{2}}$$"},{"type":"text","order":2,"title":"Apply the chain rule","content":"We now differentiate using the **chain rule**.\n\nLet the inner function be $u = 3x^5 + 5x^4$, so $y = u^{\\frac{1}{2}}$.\n\nThe chain rule states:\n\n$$\\frac{dy}{dx} = \\frac{dy}{du} \\cdot \\frac{du}{dx}$$\n\n1. Find $\\frac{dy}{du}$:\n$$\\frac{dy}{du} = \\frac{1}{2}u^{-\\frac{1}{2}}$$\n\n2. Find $\\frac{du}{dx}$:\n$$\\frac{du}{dx} = 15x^4 + 20x^3$$"},{"type":"text","order":3,"title":"Combine and simplify","content":"Substitute these parts back into the chain rule formula:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= \\frac{1}{2}(3x^5 + 5x^4)^{-\\frac{1}{2}} \\cdot (15x^4 + 20x^3) \\\\ &= \\frac{15x^4 + 20x^3}{2\\sqrt{3x^5 + 5x^4}} \\end{aligned}$$\n\nWe can factor out $x^4$ from the radical in the denominator to simplify the final fraction:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= \\frac{x^3(15x + 20)}{2\\sqrt{x^4(3x + 5)}} \\\\ &= \\frac{x^3(15x + 20)}{2x^2\\sqrt{3x+5}} \\\\ &= \\frac{x(15x+20)}{2\\sqrt{3x+5}} \\\\ &= \\frac{15x^2 + 20x}{2\\sqrt{3x+5}} \\end{aligned}$$"},{"type":"text","order":4,"title":"Verify equivalence","content":"To confirm this matches the standard product rule form from the markscheme, we can split our answer:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= \\frac{12x^2 + 20x + 3x^2}{2\\sqrt{3x+5}} \\\\ &= \\frac{4x(3x+5)}{2\\sqrt{3x+5}} + \\frac{3x^2}{2\\sqrt{3x+5}} \\\\ &= 2x\\sqrt{3x+5} + \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}} \\end{aligned}$$\n\nBoth forms are completely correct and earn full marks!"}]},{"id":"b1fe77e3-2ae4-4b96-9404-9643f868353c:logarithmic_differentiation","label":"(a) Logarithmic Differentiation","steps":[{"type":"text","order":0,"title":"Define the function","content":"To differentiate the function using **logarithmic differentiation**, we first set the function equal to $y$:\n\n$$y = x^2 \\sqrt{3x+5}$$\n\nThis method is particularly elegant for handling products and roots, as taking the natural logarithm allows us to break the expression down into simpler, individual terms."},{"type":"text","order":1,"title":"Take the logarithm","content":"Take the natural logarithm ($\\ln$) of both sides of the equation:\n\n$$\\ln(y) = \\ln\\left(x^2 \\sqrt{3x+5}\\right)$$"},{"type":"text","order":2,"title":"Simplify using log laws","content":"Using the properties of logarithms, specifically:\n- Product rule: $\\ln(ab) = \\ln(a) + \\ln(b)$\n- Power rule: $\\ln(a^n) = n \\ln(a)$\n\nFor this logarithmic method, take $x\\ne0$ and $3x+5>0$, so that $y>0$. At $x=0$, $y=0$ and logarithmic differentiation is not applicable; that case is handled separately.\n\nWe can rewrite the right-hand side. First, write the square root as a fractional power:\n\n$$\\ln(y) = \\ln\\left(x^2 (3x+5)^{\\frac{1}{2}}\\right)$$\n\nNow, expand the logarithm:\n\n$$\\ln(y) = \\ln(x^2) + \\ln\\left((3x+5)^{\\frac{1}{2}}\\right)$$\n\nSince $x\\ne0$, $\\ln(x^2)=2\\ln|x|$. Therefore,\n\n$$\\ln(y) = 2\\ln|x| + \\frac{1}{2}\\ln(3x+5)$$"},{"type":"text","order":3,"title":"Differentiate implicitly","content":"Now we differentiate both sides with respect to $x$ using the chain rule.\n\nOn the left-hand side, using implicit differentiation:\n\n$$\\frac{d}{dx}[\\ln(y)] = \\frac{1}{y} \\frac{dy}{dx}$$\n\nOn the right-hand side, we differentiate each term:\n- For $x\\ne0$, the derivative of $2\\ln|x|$ is $\\frac{2}{x}$.\n- For $\\frac{1}{2}\\ln(3x+5)$, we use the chain rule: $\\frac{1}{2} \\cdot \\frac{1}{3x+5} \\cdot 3 = \\frac{3}{2(3x+5)}$.\n\nPutting this together, we get:\n\n$$\\frac{1}{y} \\frac{dy}{dx} = \\frac{2}{x} + \\frac{3}{2(3x+5)}$$"},{"type":"text","order":4,"title":"Solve for dy/dx","content":"For $x\\ne0$, multiply both sides by $y$ to isolate the derivative $\\frac{dy}{dx}$:\n\n$$\\frac{dy}{dx} = y \\left[ \\frac{2}{x} + \\frac{3}{2(3x+5)} \\right]$$\n\nNow substitute the original expression for $y = x^2\\sqrt{3x+5}$ back into the equation:\n\n$$\\frac{dy}{dx} = x^2\\sqrt{3x+5} \\left[ \\frac{2}{x} + \\frac{3}{2(3x+5)} \\right]$$\n\nAt $x=0$, logarithmic differentiation cannot be used because $y=0$. Directly from the definition of the derivative,\n\n$$f'(0)=\\lim_{h\\to0}\\frac{h^2\\sqrt{3h+5}-0}{h}=\\lim_{h\\to0}h\\sqrt{3h+5}=0.$$"},{"type":"text","order":5,"title":"Simplify the expression","content":"For $x\\ne0$, distribute the $x^2\\sqrt{3x+5}$ terms:\n\n$$\\frac{dy}{dx} = \\frac{2x^2\\sqrt{3x+5}}{x} + \\frac{3x^2\\sqrt{3x+5}}{2(3x+5)}$$\n\nSimplify each term:\n- For the first term: $\\frac{2x^2\\sqrt{3x+5}}{x} = 2x\\sqrt{3x+5} = 2x(3x+5)^{\\frac{1}{2}}$\n- For the second term, write the denominator as $(3x+5)^1$ and use the laws of indices: $\\frac{(3x+5)^{\\frac{1}{2}}}{(3x+5)^1} = (3x+5)^{-\\frac{1}{2}}$:\n\n$$\\frac{3x^2\\sqrt{3x+5}}{2(3x+5)} = \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}}$$\n\nThus, for $x\\ne0$, we get:\n\n$$\\frac{dy}{dx} = 2x(3x+5)^{\\frac{1}{2}} + \\frac{3}{2}x^2(3x+5)^{-\\frac{1}{2}}$$\n\nAt $x=0$, the derivative is $0$, and the same final expression also evaluates to $0$. Hence this is the correct derivative wherever the derivative exists."}]},{"id":"58d96fce-5e03-4c41-b54e-2b57b7f96934:product_chain_rule","label":"(b) Product and Chain Rules","steps":[{"type":"text","order":0,"title":"Identify the product rule","content":"We need to differentiate the function $y = x \\ln(4x-1)$. Since this function is the product of two simpler functions, we will use the **Product Rule**: $$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$ Let's define our two parts: $u = x$ and $v = \\ln(4x-1)$."},{"type":"text","order":1,"title":"Differentiate each part","content":"Now we differentiate each part with respect to $x$. For $u = x$, we have $$\\frac{du}{dx} = 1$$ For $v = \\ln(4x-1)$, we use the **Chain Rule**: $$\\frac{dv}{dx} = \\frac{1}{4x-1} \\cdot \\frac{d}{dx}(4x-1) = \\frac{4}{4x-1}$$ This earns the first method mark **M1**."},{"type":"text","order":2,"title":"Apply product rule formula","content":"Substitute our components into the Product Rule formula: $$\\frac{dy}{dx} = u\\frac{dv}{dx} + v\\frac{du}{dx}$$ This gives: $$\\frac{dy}{dx} = x \\cdot \\left(\\frac{4}{4x-1}\\right) + \\ln(4x-1) \\cdot 1$$ This earns the second method mark **M1**."},{"type":"text","order":3,"title":"Simplify the final expression","content":"Simplify the expression to write it in its cleanest form: $$\\frac{dy}{dx} = \\ln(4x-1) + \\frac{4x}{4x-1}$$ This earns the final accuracy mark 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Differentiation (A-level only)","topicId":"07681e27-ffd1-457b-97e1-6d9f7376045a","topicSidebar":"$L3b","topicSlug":"differentiation-9","topicTitle":"3.5 Differentiation (A-level only)"}]}],"$L3c","$L3d"]}]

3.5 Differentiation (A-level only)

Question 63

3.5 Differentiation (A-level only) Questions

3.5 Differentiation (A-level only)

Question 63

3.5 Differentiation (A-level only) Questions