Further Differentiation (Cambridge (CIE) AS Maths: Pure 2): Exam Questions

(i) Differentiate 4 cos x - 3 sin x

(ii) Use the product rule to differentiate e^x In x

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

(iv) Use the chain rule to differentiate cos(x² - 7x + 1)

A curve has the equation y = e ^-3x + Inx, 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 $\frac{\mathrm{dy}}{\mathrm{dx}}$for each of the following:

y = cos(x² - 3x + 7) + sin(e^x)

y = In (2x³)

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 ^3x-5

(x³ - 4x² + 7) In x

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

Show that if y = cosec 2x, then

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

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

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

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

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

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

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

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

Use an appropriate method to differentiate each of the following.

(i) sin 2x - e^7x

(ii) x² In x

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

(iv) In(tan x)

A curve has the equation y = e^-3x + In 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 = In(axⁿ), where a > 0 is a real number and n ≥ 1 is an integer, show that

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

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

(2 sin 3х - cos 3x)e^6-×

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

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

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

Given that x = sec 7y,

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of y

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

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

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

Point A is a maximum point on the graph.

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

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

Use an appropriate method to differentiate each of the following.

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

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

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

(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 $\left(1,\frac{7}{2}\right)\mathrm{is}$

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

Show that the derivative y = 4^-×4 is

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

Hence find the equation of the tangent to the curve at the point (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:

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

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

The diagram below shows the graph of y = f(x), where f(x) is the function defined by

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

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

Find the range of f(x), giving your answer correct to 3 decimal places.