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

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 an appropriate method to differentiate each of the following.

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

(ii) $x^2\text{ln}x$

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

(iv) $\text{ln}\left(\text{tan}x\right)$

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

Differentiate $\left(2\text{sin}3x-\text{cos}3x\right){\text{e}}^{6-x}$ with respect to $x$, simplifying your answer as far as possible.

Differentiate ${\left(x^2-x\right)}^2\text{ln}5x$ with respect to $x$, simplifying your answer as far as possible.

Use an appropriate method to differentiate each of the following.

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

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

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

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

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

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

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

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

$\frac{\text{d}y}{\text{d}x}=\frac{n}{x}$

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

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

Given that $x=\text{sec}7y$, find $\frac{\text{d}y}{\text{d}x}$ in terms of $y$.

Hence find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$.

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

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

Point $A$ is a maximum point on the graph.

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

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

A curve has 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

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

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

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

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

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 $\left(5+{\text{sin}}^{2}3x\right){\text{e}}^{x^2-3x+2}$ with respect to $x$, simplifying your answer where possible.

Differentiate $3^{\sqrt{x}}\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)$ with respect to $x$, simplifying your answer where possible.

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

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

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

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