Further Differentiation (Edexcel International A Level (IAL) Maths: Pure 3): Exam Questions

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

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

Find  $\frac{dy}{dx}$  for each of the following:

$y=\cos \left(x^2-3x+7\right)+\sin \left(e^x\right)$

Find $\frac{dy}{dx}$  for each of the following:

$y=\ln \left(2x^3\right)$

Find the equation of the tangent to the curve  $y=e^{3x^2+5x-2}$  at the point $\left(-2,1\right)$, 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:

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

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

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

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

$\frac{dy}{dx}=-2\mathrm{cosec}2x\cot 2x$

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

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

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

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

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

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

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

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

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

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 difference between the y-values of the coordinates of A and B, giving your answer correct to 3 decimal places.

A is the point on the graph of  $y=\arctan x$such that the tangent to the graph at $A$ passes through the point $\left(0,\frac{1}{2}\right)$.  Show that the x-coordinate of A satisfies the equation

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

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