Differentiation: chain, product and quotient rules – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Mark scheme: 2 marks for each part.

   1. Use \(\dfrac{d}{dx}\ln u=\dfrac{u'}{u}\) with \(u=4x^2+1\), so \(u'=8x\). \(\dfrac{dy}{dx}=\dfrac{8x}{4x^2+1}\) (1 for the chain rule, 1 for simplifying). So \(\dfrac{dy}{dx}=\dfrac{8x}{4x^2+1}\).
   2. Take \(u=x^3\), \(v=e^{-2x}\); the chain rule gives \(v'=-2e^{-2x}\). \(\dfrac{dy}{dx}=3x^2e^{-2x}-2x^3e^{-2x}=x^2e^{-2x}(3-2x)\) (1 for each term, then factorising). So \(a=3\) and \(b=-2\).

   Common errors: in (a), writing \(\dfrac1{4x^2+1}\) without multiplying by the derivative of the inside; in (b), forgetting the factor \(-2\) from \(e^{-2x}\), which gives the wrong bracket \((3-x)\). Check (b) at \(x=1\): \(e^{-2}(3-2)=e^{-2}\approx0.135\).

2. Answer to Question 2 (4 marks)

   Mark scheme: 2 marks for the derivative, 1 mark for each stationary point.

   1. Use the quotient rule with \(u=x^2\) and \(v=2x+1\). \(\dfrac{dy}{dx}=\dfrac{(2x+1)2x-x^2\cdot2}{(2x+1)^2}=\dfrac{2x^2+2x}{(2x+1)^2}=\dfrac{2x(x+1)}{(2x+1)^2}\) (1 for the method, 1 for the result). This is the required result.
   2. A stationary point needs a zero numerator: \(2x(x+1)=0\), so \(x=0\) or \(x=-1\). \(x=0\): \((0,0)\). \(x=-1\): \(y=\dfrac{1}{-1}=-1\): \((-1,-1)\). (1 mark each.) So the stationary points are \((0,0)\) and \((-1,-1)\).

3. Answer to Question 3 (4 marks)

   Mark scheme: 1 mark for checking the point, 1 mark for the implicit differentiation, 1 mark for the gradient, 1 mark for the tangent.

   1. Substitute the point, then differentiate each term; \(2xy\) needs the product rule. \(9+6+3=18\) so \((3,1)\) lies on the curve (1). \(2x+2y+2x\dfrac{dy}{dx}+6y\dfrac{dy}{dx}=0\) (1), so \(\dfrac{dy}{dx}=-\dfrac{x+y}{x+3y}=-\dfrac46=-\dfrac23\) (1). So the gradient at \((3,1)\) is \(-\tfrac23\).
   2. Use \(y-y_1=m(x-x_1)\) and multiply by 3. \(y-1=-\dfrac23(x-3)\), i.e. \(2x+3y=9\) (1). So the tangent is \(2x+3y=9\).

4. Answer to Question 4 (4 marks)

   Mark scheme: 2 marks for \(\dfrac{dy}{dx}\); 1 mark for the point and gradient, 1 mark for the normal.

   1. Find \(\dfrac{dx}{dt}=2t\) and \(\dfrac{dy}{dt}=6t^2-1\), then divide. \(\dfrac{dy}{dx}=\dfrac{6t^2-1}{2t}\) (2). This is undefined at \(t=0\), where the tangent is vertical.
   2. Substitute \(t=1\): \(x=3\), \(y=1\). The normal gradient is the negative reciprocal of \(\tfrac52\). At \(t=1\) the point is \((3,1)\) and the gradient is \(\dfrac52\) (1). The normal is \(y-1=-\dfrac25(x-3)\), i.e. \(2x+5y=11\) (1). So the normal is \(2x+5y=11\).

5. Answer to Question 5 (4 marks)

   Method: link the rates by the chain rule. First find \(\dfrac{dr}{dt}\) from the known rate of change of volume, then use it to find the rate of change of area.

   \(\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}\) (1). \(100=100\pi\dfrac{dr}{dt}\), so \(\dfrac{dr}{dt}=\dfrac1\pi=0.318\) cm s\(^{-1}\) (1). \(\dfrac{dA}{dt}=8\pi r\dfrac{dr}{dt}\) (1) \(=8\pi\times5\times\dfrac1\pi=40\) cm\(^2\) s\(^{-1}\) (1).

   Answer: when \(r=5\) cm, the radius increases at about 0.318 cm s\(^{-1}\) and the surface area at 40 cm² s\(^{-1}\).

Total : 20 marks