Further differentiation (chain, product, quotient, implicit, parametric, connected rates) – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ when

(a) $y=(3x^{2}-1)^{5}$ [2]

(b) $y=x^{2}\mathrm{e}^{3x}$ [2]

(a) Chain rule: $k x(3x^{2}-1)^{4}$M1
$30x(3x^{2}-1)^{4}$A1
(b) Product rule: $Ax\mathrm{e}^{3x}+Bx^{2}\mathrm{e}^{3x}$M1
$2x\mathrm{e}^{3x}+3x^{2}\mathrm{e}^{3x}$ (or $x\mathrm{e}^{3x}(2+3x)$)A1
Answer: (a) $30x(3x^{2}-1)^{4}$ (b) $x\mathrm{e}^{3x}(2+3x)$
Examiner tip: In (a) don't forget to multiply by the derivative of the inside, $6x$.

2StandardCalculator5 marks

The curve $C$ has equation $$y=\frac{\ln x}{x^{2}},\quad x>0$$

(a) Show that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1-2\ln x}{x^{3}}$. [3]

(b) Find the exact coordinates of the stationary point of $C$. [2]

(a) Quotient rule (or product rule on $x^{-2}\ln x$): $\dfrac{x^{2}\cdot\frac1x-\ln x\cdot2x}{(x^{2})^{2}}$M1
Correct unsimplified derivative, e.g. $\dfrac{x-2x\ln x}{x^{4}}$A1
Divides through by $x$ to reach $\dfrac{1-2\ln x}{x^{3}}$ csoA1*
(b) Sets $1-2\ln x=0\Rightarrow x=\mathrm{e}^{\frac12}$M1
$\left(\sqrt{\mathrm{e}},\ \dfrac{1}{2\mathrm{e}}\right)$A1
Answer: (b) $\left(\mathrm{e}^{\frac12},\ \dfrac{1}{2\mathrm{e}}\right)$
Examiner tip: 'Exact' means leave $\mathrm{e}^{1/2}$ and $\tfrac{1}{2\mathrm{e}}$ – decimals lose the final A mark.

3StandardCalculator7 marks

The curve $C$ has equation $$x^{2}+3xy+y^{2}=11$$

(a) Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$. [4]

(b) Find an equation of the tangent to $C$ at the point $(1,2)$, giving your answer in the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers. [3]

(a) Implicit differentiation: $y^{2}\to2y\dfrac{\mathrm{d}y}{\mathrm{d}x}$ (and $x^2\to2x$, $11\to0$)M1
$3xy\to3y+3x\dfrac{\mathrm{d}y}{\mathrm{d}x}$B1
Collects $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ terms and factorisesM1
$\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{2x+3y}{3x+2y}$A1
(b) Substitutes $(1,2)$ to get gradient $-\tfrac87$M1
Uses $y-2=m(x-1)$ with their gradientM1
$8x+7y-22=0$ (or any integer multiple)A1
Answer: (a) $-\dfrac{2x+3y}{3x+2y}$ (b) $8x+7y-22=0$
Examiner tip: $3xy$ is a product – differentiating it as $3y\frac{\mathrm{d}y}{\mathrm{d}x}$ is the most common way to lose the B1.

4StandardCalculator9 marks

A curve $C$ has parametric equations $$x=t^{2}+1,\qquad y=t^{3}-3t,\qquad t\in\mathbb{R}$$

(a) Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $t$. [2]

(b) Find the coordinates of the points on $C$ at which the tangent is parallel to the $x$-axis. [3]

(c) Find an equation of the normal to $C$ at the point where $t=2$, giving your answer in the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers. [4]

(a) Finds $\dfrac{\mathrm{d}x}{\mathrm{d}t}=2t$, $\dfrac{\mathrm{d}y}{\mathrm{d}t}=3t^{2}-3$ and divides the right way roundM1
$\dfrac{3t^{2}-3}{2t}$A1
(b) Sets $3t^{2}-3=0\Rightarrow t=\pm1$M1
$(2,-2)$A1
$(2,2)$A1
(c) Point $(5,2)$B1
Gradient of tangent $\tfrac94$, uses normal gradient $-\dfrac{1}{m}=-\tfrac49$M1
$y-2=-\tfrac49(x-5)$ with their point and normal gradientM1
$4x+9y-38=0$ (or any integer multiple)A1
Answer: (a) $\dfrac{3t^{2}-3}{2t}$ (b) $(2,-2)$ and $(2,2)$ (c) $4x+9y-38=0$
Examiner tip: Convert $t$ values back to $(x,y)$ coordinates – giving $t=\pm1$ as the answer to (b) loses both A marks.

5HardCalculator6 marks

A spherical balloon is being inflated so that its volume increases at a constant rate of $50\ \text{cm}^{3}\,\text{s}^{-1}$.

[The volume of a sphere is $V=\tfrac43\pi r^{3}$ and its surface area is $S=4\pi r^{2}$.]

(a) Find the rate of increase of the radius at the instant when the radius is 10 cm, giving your answer to 3 significant figures. [3]

(b) Find the rate of increase of the surface area at this instant. [3]

(a) $\dfrac{\mathrm{d}V}{\mathrm{d}r}=4\pi r^{2}$B1
Uses $\dfrac{\mathrm{d}r}{\mathrm{d}t}=\dfrac{\mathrm{d}V}{\mathrm{d}t}\div\dfrac{\mathrm{d}V}{\mathrm{d}r}=\dfrac{50}{4\pi(10)^{2}}$M1
$0.0398\ \text{cm}\,\text{s}^{-1}$ (accept $\tfrac{1}{8\pi}$)A1
(b) $\dfrac{\mathrm{d}S}{\mathrm{d}r}=8\pi r$B1
$\dfrac{\mathrm{d}S}{\mathrm{d}t}=\dfrac{\mathrm{d}S}{\mathrm{d}r}\times\dfrac{\mathrm{d}r}{\mathrm{d}t}=80\pi\times\dfrac{1}{8\pi}$M1
$10\ \text{cm}^{2}\,\text{s}^{-1}$ (accept awrt $10.0$ from using a rounded $\frac{\mathrm{d}r}{\mathrm{d}t}$)A1
Answer: (a) $0.0398\ \text{cm s}^{-1}$ (b) $10\ \text{cm}^{2}\text{ s}^{-1}$
Examiner tip: Write the chain rule out with all three derivatives before substituting – it shows the method even if a number slips.

6HardCalculator5 marks

Given that $x$ is measured in radians, use differentiation from first principles to show that $$\frac{\mathrm{d}}{\mathrm{d}x}(\cos x)=-\sin x$$

You may use, without proof, the results that as $h\to0$, $\dfrac{\sin h}{h}\to1$ and $\dfrac{\cos h-1}{h}\to0$. [5]

Writes the gradient of a chord as $\dfrac{\cos(x+h)-\cos x}{h}$ (allow $\delta x$ in place of $h$)B1
Uses $\cos(x+h)=\cos x\cos h-\sin x\sin h$M1
Groups terms to obtain $\cos x\left(\dfrac{\cos h-1}{h}\right)-\sin x\left(\dfrac{\sin h}{h}\right)$M1
Applies both given limits correctly as $h\to0$: $\cos x\times0-\sin x\times1$A1
Complete proof with correct limit notation throughout, concluding $\dfrac{\mathrm{d}}{\mathrm{d}x}(\cos x)=\lim_{h\to0}\left[\dots\right]=-\sin x$ csoA1*
Answer: Proof: $\displaystyle\lim_{h\to0}\left[\cos x\left(\frac{\cos h-1}{h}\right)-\sin x\left(\frac{\sin h}{h}\right)\right]=-\sin x$
Examiner tip: Keep writing '$\lim_{h\to0}$' until you actually take the limit – dropping it, or putting $h=0$ early, costs the final mark.

7StretchCalculator9 marks

The curve $C$ has equation $$y=\mathrm{e}^{-x}\sin x,\qquad 0\leqslant x\leqslant2\pi$$

(a) Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$. [2]

(b) Find the exact $x$-coordinates of the stationary points of $C$. [2]

(c) Show that $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=-2\mathrm{e}^{-x}\cos x$ and hence determine the nature of each stationary point. [3]

(d) Find the set of values of $x$ for which $C$ is concave. [2]

(a) Product rule: $\pm\mathrm{e}^{-x}\sin x\pm\mathrm{e}^{-x}\cos x$M1
$\mathrm{e}^{-x}(\cos x-\sin x)$A1
(b) $\cos x-\sin x=0\Rightarrow\tan x=1$ (with $\mathrm{e}^{-x}\neq0$)M1
$x=\dfrac{\pi}{4},\ \dfrac{5\pi}{4}$A1
(c) Product rule on $\mathrm{e}^{-x}(\cos x-\sin x)$: $-\mathrm{e}^{-x}(\cos x-\sin x)+\mathrm{e}^{-x}(-\sin x-\cos x)$M1
Simplifies to $-2\mathrm{e}^{-x}\cos x$A1*
At $x=\tfrac{\pi}{4}$, $y''<0$ so maximum; at $x=\tfrac{5\pi}{4}$, $y''>0$ so minimum (both with reasons)A1
(d) Concave where $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}<0\Rightarrow\cos x>0$M1
$0\leqslant x<\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}A1
Answer: (b) $\tfrac{\pi}{4},\tfrac{5\pi}{4}$ (c) max at $\tfrac{\pi}{4}$, min at $\tfrac{5\pi}{4}$ (d) $0\leqslant x<\tfrac{\pi}{2}$, $\tfrac{3\pi}{2}
Examiner tip: Concave means $f''(x)<0$ (curve bends downwards) – mixing up concave and convex throws away part (d).

8StandardCalculator5 marks

(a) Find the gradient of the curve $y=5^{x}$ at the point where $x=2$, giving your answer in the form $k\ln5$. [2]

(b) Given that $x=\tan y$, $-\dfrac{\pi}{2}[3]

(a) $\dfrac{\mathrm{d}y}{\mathrm{d}x}=5^{x}\ln5$M1
$25\ln5$A1
(b) $\dfrac{\mathrm{d}x}{\mathrm{d}y}=\sec^{2}y$M1
Uses $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{\mathrm{d}x/\mathrm{d}y}$ and $\sec^{2}y=1+\tan^{2}y$M1
$\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{1+\tan^{2}y}=\dfrac{1}{1+x^{2}}$ csoA1*
Answer: (a) $25\ln5$
Examiner tip: The final answer in (b) must be in terms of $x$ – you must use the identity $\sec^{2}y=1+\tan^{2}y$, not just stop at $\cos^{2}y$.

Still losing marks on this topic?

Book a free introductory session and we'll find exactly which step is costing you.

Book a free intro