Further Differentiation (Cambridge (CIE) A Level Maths: Pure 3): Exam Questions

(i) Differentiate $4\text{cos}x-3\text{sin}x$

(ii) Use the product rule to differentiate ${\text{e}}^x\text{ln}x$

(iii) Use the quotient rule to differentiate $\frac{\text{tan}x}{x}$

(iv) Use the chain rule to differentiate $\text{cos}\left(x^2-7x+1\right)$

A curve has equation $y={\text{e}}^{-3x}+\text{ln}x$, $x>0$.

Find the gradient of the normal to the curve at the point $\left(1,{\text{e}}^{-3}\right)$, giving your answer correct to 3 decimal places.

Find $\frac{\text{d}y}{\text{d}x}$ for $y=\text{cos}\left(x^2-3x+7\right)+\text{sin}\left({\text{e}}^x\right)$.

Find $\frac{\text{d}y}{\text{d}x}$ for $y=\text{ln}\left(2x^3\right)$.

Find the equation of the tangent to the curve $y={\text{e}}^{3x^2+5x-2}$ at the point $(-2,1)$, giving your answer in the form $ax+by+c=0$, where $a$, $b$ and $c$ are integers.

Differentiate with respect to $x$, simplifying your answers as far as possible:

$\left(4\text{cos}x-3\text{sin}x\right){\text{e}}^{3x-5}$

$\left(x^3-4x^2+7\right)\text{ln}x$

Differentiate $\frac{5x^7}{\text{sin}2x}$ with respect to $x$.

Show that if $y=\text{cosec}2x$, then

$\frac{\text{d}y}{\text{d}x}=-2\text{cosec}2x\text{cot}2x$

Hence find the gradient of the tangent to the curve $y=\text{cosec}2x$ at the point with coordinates $\left(\frac{\pi }{3},\frac{2\sqrt{3}}{3}\right)$.

The diagram below shows part of the graph of $y=\text{f}\left(x\right)$, where $\text{f}\left(x\right)$ is the function defined by

$\text{f}\left(x\right)=\left(x^2-1\right)\text{ln}(x+3),x>-3$

Points $A$, $B$ and $C$ are the three places where the graph meets the $x$-axis.

Find $\text{f}\text{'}\left(x\right)$.

Show that the coordinates of point $A$ are $(-2,0)$.

Find the equation of the tangent to the curve at point $A$.

Use the chain rule to differentiate ${\text{tan}}^{-1}\left(ax\right)$, where $a$ is a real constant.

What does your answer to part (a) tell you about the number of stationary points on the curve $y={\text{tan}}^{-1}\left(3x\right)$?

Use an appropriate method to differentiate each of the following.

i) $\sin 2x-e^{7x}$

ii) $x^2\ln x$

iii) $\frac{\cos 3x}{\tan 2x}$

iv) $\ln \left(\tan x\right)$

A curve has the equation  $y=e^{-3x}+\ln x,x>0.$

Show that the equation of the tangent to the curve at the point with x-coordinate 1 is

$y=\left(\frac{e^3-3}{e^3}\right)x+\frac{4-e^3}{e^3}$

For $y=\ln \left(a{x}^n\right)$ , where $a>0$ is a real number and $n\ge 1$is an integer, show that

$\frac{dy}{dx}=\frac{n}{x}$

Find the gradient of the normal to the curve $y=5\cos (e^x-\frac{\pi }{2})$at the point with x-coordinate 0.  Give your answer correct to 3 decimal places. 

Differentiate with respect to x, simplifying your answers as far as possible:

$\left(2\sin 3x-\cos 3x\right)e^{6-x}$

${\left(x^2-x\right)}^2\ln 5x$

By writing  $y=\frac{\mathrm{f}\left(x\right)}{\mathrm{g}\left(x\right)}$  as $y={\mathrm{f}\left(x\right)\left[\mathrm{g}\left(\mathrm{x}\right)\right]}^{-1}$ and then using the product and chain rules, show that

$\frac{dy}{dx}=\frac{\mathrm{g}\left(x\right)\mathrm{f}\text{'}\left(x\right)-\mathrm{f}\left(x\right)\mathrm{g}\text{'}\left(x\right)}{{\left(g\left(x\right)\right)}^2}$

Given that $x=\sec 7y$ ,

Find  $\frac{dy}{dx}$  in terms of y

Hence find $\frac{dy}{dx}$ in terms of x.

The diagram below shows part of the graph of $y=\mathrm{f}\left(x\right)$, where $\mathrm{f}\left(x\right)$is the function defined by

$\mathrm{f}\left(x\right)=\frac{\sin x}{1-e^x},x>0$

Point A is a maximum point on the graph.

Show that the x-coordinate of A is a solution to the equation

$\frac{\cos x+e^x\left(\sin x-\cos x\right)}{e^{2x}-2e^x+1}=0$

Use the chain rule to show that the derivative of $y={\tan }^{-1}\left(\frac{x}{a}\right)$, where $a\ne 0$ is a real constant, is a

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{a}{a^2+x^2}$

Hence find the coordinates of any stationary point(s) on the curve with equation

$y=-\frac{x}{4}+{\tan }^{-1}\left(\frac{x}{2}\right)$

giving your answers as exact values.

Use an appropriate method to differentiate each of the following.

i) $\tan 3x+e^{7-2x^2}$

ii) $\left(x^2+2x-8\right)\cos \left(3-x\right)$

iii) $\frac{\ln 7x}{\sin \left(x^2+5\right)}$

iv) $\sqrt{\cos 4x}$

A curve has the equation $y=3^x+2^{-x}$.

Show that the gradient of the normal to the curve at the point  $\left(1,\frac{7}{2}\right)$ is

$\frac{2}{\ln 2-6\ln 3}$

Find the derivative of the function $\mathrm{f}\left(x\right)=\sin \left(\cos \left(\ln \frac{1}{x}\right)\right),x>0.$

Show that the derivative  $y=4^{-x^4}$  is

$\frac{dy}{dx}=-\left(\ln 4\right)x^3{4}^{1-x^4}$

Hence find the equation of the tangent to the curve at the point $\left(1,\frac{1}{4}\right)$, giving your answer in the form $y=ax+b$, where a and b are to be given as exact values.

Differentiate with respect to x, simplifying your answers where possible:

$\left(5+{\sin }^{2}3x\right)e^{x^2-3x+2}$

$3^{\sqrt{x}}\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)$

The diagram below shows the graph of $y=\mathrm{f}\left(x\right)$, where $\mathrm{f}\left(x\right)$ is the function defined by

$\mathrm{f}\left(x\right)=\frac{\sin 3x}{e^{2x-3}},0\le x\le \frac{2\pi }{3}$

The points A and B are maximum and minimum points, respectively.

Find the range of $\mathrm{f}\left(x\right)$ , giving your answer correct to 3 decimal places.

A sequence of functions is defined by the recurrence relation

$u_{k+1}\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}{u}_k\left(x\right),u_1\left(x\right)=\sin \left(x\sqrt{2}\right)$

Based on that sequence, the function${\mathrm{f}}_n\left(x\right)$is defined by

${\mathrm{f}}_n\left(x\right)=\sum _{r=1}^n{u}_r\left(x\right)$

Calculate the value of ${\mathrm{f}}_{41}\left(\frac{\mathrm{\pi }\sqrt{2}}{4}\right)$

Use calculus to find the coordinates of the stationary points of the curve

$y=\frac{x}{3}-{\tan }^{-1}\left(\frac{2x}{3}\right)$

and determine whether each one is a maximum or a minimum.  The coordinates should be given as exact values.