r":"sky","style":"text-color","anchor":"term1"},{"color":"purple","style":"text-color","anchor":"term2"}],"connections":[{"to":"term1","from":"du-sub","color":"sky","label":"u' v","style":"solid"},{"to":"term2","from":"dv-sub","color":"purple","label":"u v'","style":"solid"}]},"order":2,"title":"Apply the product rule","description":"Combine the derivatives using the product rule formula: $u'v + uv'$."},{"type":"text","order":3,"title":"Simplify the expression","content":"Finally, simplify the terms to obtain our final answer:\n\n$$\\frac{dy}{dx} = \\ln(2x+1) + \\frac{2x}{2x+1}$$\n\n**Exam Tip on Mark Allocation:**\n* **M1** is awarded for correctly differentiating $\\ln(2x+1)$ to get $\\frac{2}{2x+1}$ using the chain rule.\n* **M1** is awarded for applying the product rule formula correctly.\n* **A1** is awarded for writing the final simplified expression correctly."}]}],"generatedAt":"2026-06-09T09:17:25.343Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"fcc4935a-6dc7-4c9c-afb0-ba70d84f23a6","revisionId":"a74b0f74-c3c7-4d25-98c9-3199e2809828","questionNumber":null,"questionType":"LA","specification":"The point $P$ lies on the curve with equation $y = (4x - 3) \\ln(x - 1)$ with $x$ coordinate 2.","difficulty":2,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"7f79b34b-1c09-4662-9d91-340b1d2ed4e6","label":"a","content":"Find an equation of the tangent to the curve at the point $P$.","markscheme":"- Differentiates using product rule: $\\frac{dy}{dx} = 4 \\ln(x - 1) + \\frac{4x - 3}{x - 1}$ M1 A1\n- Substitutes $x = 2$ into $\\frac{dy}{dx}$ to find gradient $m = 5$ M1\n- Finds $y$-coordinate at $x = 2$: $y = (4(2) - 3) \\ln(2 - 1) = 0$ B1\n- Uses $y - y_1 = m(x - x_1)$ to find $y = 5x - 10$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Attempt to differentiate using the product rule.","code":"M"},{"id":"p2","marks":1,"description":"Correct derivative expression $\\frac{dy}{dx} = 4 \\ln(x - 1) + \\frac{4x - 3}{x - 1}$.","code":"A"},{"id":"p3","marks":1,"description":"Correct substitution of $x = 2$ to find the gradient $m = 5$.","code":"M"},{"id":"p4","marks":1,"description":"Correct evaluation of $y$-coordinate at $x = 2$, $y = 0$.","code":"B"},{"id":"p5","marks":1,"description":"Correct equation of the tangent, e.g., $y = 5x - 10$.","code":"A"}],"finalAnswers":["y = 5x - 10","y - 0 = 5(x - 2)"],"generatedAt":"2026-05-06T14:03:57.404Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"calculus_analytic","label":"Standard Calculus Approach","steps":[{"type":"text","order":0,"title":"Find the y-coordinate","content":"To find the coordinates of point $P$, substitute the given $x$-coordinate, $x = 2$, into the original curve equation:\n\n$$y = (4x - 3) \nln(x - 1)$$\n\nSubstituting $x = 2$:\n\n$$\\begin{aligned} y &= (4(2) - 3) \\ln(2 - 1) \\\\ y &= (8 - 3) \\ln(1) \\\\ y &= 5 \\times 0 \\\\ y &= 0 \\end{aligned}$$\n\nSo, the coordinates of point $P$ are $(2, 0)$. This earns us the first mark.","highlights":[{"text":"y = (4x - 3) \\ln(x - 1)","location":"specification"},{"text":"x coordinate 2","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"To find the gradient of the tangent, we first differentiate $y$ with respect to $x$. Since the function is a product of two terms, we use the product rule:"},{"latex":"$$$\\frac{d}{dx}(\\htmlData{anchor=u-term}{\\color{@sky}{u}}\\htmlData{anchor=v-term}{\\color{@amber}{v}}) = \\htmlData{anchor=u-deriv}{\\color{@sky}{u'}}\\htmlData{anchor=v-orig}{\\color{@amber}{v}} + \\htmlData{anchor=u-orig}{\\color{@sky}{u}}\\htmlData{anchor=v-deriv}{\\color{@amber}{v'}}$$"},{"text":"Let $\\color{@sky}{u = 4x - 3}$ and $\\color{@amber}{v = \\ln(x - 1)}$. Finding their derivatives:"},{"latex":"$$$\\htmlData{anchor=u-prime}{\\color{@sky}{u'}} = 4 \\quad \\text{and} \\quad \\htmlData{anchor=v-prime}{\\color{@amber}{v'}} = \\f

Differentiation

Question 30

Differentiation Questions

Differentiation

Question 30

Differentiation Questions