Further Differentiation (Edexcel IAL Maths): Exam Questions

A curve has the equation $y=5e^{-2x}$.

Find an expression for $\frac{dy}{dx}$.

(i) Find the gradient of the tangent at the point where $x=1$, giving your answer in the form $-ae^{-2}$ where a is a positive integer to be found.

(ii) Hence show that the gradient of the normal to the curve at the point where $x=1$ is $\frac{1}{10}e^{2}$.

Find $\frac{dy}{dx}$for

(i) $y = \sin ( 3 x^{2} )$,

(ii) $y = 2 \ln ( x^{3} )$ .

The curve with equation $y=e^{x^{2}-9}$ passes through the point with coordinates (-3 , 1).

(i) Find an expression for $\frac{dy}{dx}$.

(ii) Find the equation of the tangent to the curve at the point (-3 , 1).

Differentiate $(x^{3}-2x)\mathrm{ln}x$with respect to x.

Differentiate $e^{x}\mathrm{cos}2x$ with respect to x.

Differentiate $\frac{\mathrm{cos}x}{\mathrm{sin}x}$ with respect to x

Differentiate $\frac{2x^{2}-3x+4}{\mathrm{sin}3x}$ with respect to $x$.

Write down $\frac{dy}{dx}$ when

(i) $y=\mathrm{sec}5x,$

(ii) $y=\mathrm{cosec}3x.$

The function $f(x)$ is defined as

$\mathrm{f} ( x ) = ( x^{2} - 4 x + 4 ) \ln ( x ) , x > 0$

Show that the graph of $y = \mathrm{f} ( x )$ intercepts the x-axis at the points (1 , 0) and (2 , 0).

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

Find the gradient of the tangent at the point (1 , 0).

Hence find the equation of the tangent at the point (1 , 0), giving your answer in the form $ax+by+c=0$, where a, b and c are integers to be found.

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

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

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

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

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

$y = \ln ( 2 x^{3} )$

Find the equation of the tangent to the curve $y=e^{3x^{2}+5x-2}$ 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:

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

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

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

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

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

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

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

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

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

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

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 $y=e^{-3x}+\mathrm{ln}x,x>0.$

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

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

For $y = \ln ( a x^{n} )$ , where $a>0$ is a real number and $n\geq1$is an integer, show that

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

Find the gradient of the normal to the curve $y=5\mathrm{cos}(e^{x}-\frac{π}{2})$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:

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

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

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

$\frac{d y}{d x} = \frac{ \mathrm{g} ( x ) \mathrm{f} ' ( x ) - \mathrm{f} ( x ) \mathrm{g} ' ( x )}{( g ( x ) )^{2}}$

Given that $x=\mathrm{sec}7y$ ,

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 $y = \mathrm{f} ( x )$, where $\mathrm{f} ( x )$is the function defined by

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

Point A is a maximum point on the graph.

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

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

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

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

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

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

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

$\frac{d y}{d x} = - ( \ln 4 ) x^{3} 4^{1 - x^{4} }$

Hence find the equation of the tangent to the curve at the point $( 1 , 1 \frac{1}{4} )$, 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:

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

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

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

$\mathrm{f} ( x ) = \frac{\sin 3 x}{e^{2 x - 3} } ,   0 \text{less} \text{or} \text{equal} \text{than} x \text{less} \text{or} \text{equal} \text{than} \frac{2 \pi }{3}$

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

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

A sequence of functions is defined by the recurrence relation

$u_{k + 1} ( x ) = \frac{ \mathrm{d} }{ \mathrm{d} x} u_{k} ( x ) , u_{1} ( x ) = \sin ( x \sqrt{2} )$

Based on that sequence, the function$\mathrm{f}_{n} ( x )$is defined by

$\mathrm{f}_{n} ( x ) = \sum_{r = 1}^{n} u_{r} ( x )$

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