Further Differentiation (AQA A Level Maths: Pure): Exam Questions

Figure 1 shows a sketch of part of the curve $C$ with equation $y=\text{f}\left(x\right)$, where

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

The curve $C$ crosses the $x$-axis at the points $A$, $B$ and $C$, as shown in Figure 1.

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

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

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

Given that

$y=\ln \left(a{x}^n\right)$

where $a>0$ is a real constant and $n⩾1$ is an integer,

show that

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

Given that $\text{f}\left(x\right)=\text{sin}x$, where $x$ is measured in radians,

use differentiation from first principles to show that

$\text{f}\text{'}\left(x\right)=\underset{h\to 0}{\lim }\left(\text{sin}x\left(\frac{\text{cos}h-1}{h}\right)+\text{cos}x\left(\frac{\text{sin}h}{h}\right)\right)$

Hence prove that $\text{f}\text{'}\left(x\right)=\text{cos}x$.

The curve $C$ has equation $y=\text{f}\left(x\right)$, where

$\text{f}\left(x\right)=(x^2-4x+4)\ln xx>0$

Show that $C$ meets the $x$-axis at the points $(1,0)$ and $(2,0)$.

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

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

Hence find the equation of the tangent to $C$ 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 $C$ has equation

$y={\text{e}}^{-3x}+\ln xx>0$

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

Given that

$y=\frac{\text{f}\left(x\right)}{\text{g}\left(x\right)}$

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

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

Given that

$y=\text{cos}(x^2-3x+7)+\text{sin}\left({\text{e}}^x\right)x\in \mathrm{ℝ}$

find $\frac{\text{d}y}{\text{d}x}$.

Given that

$y=\ln \left(2x^3\right)x>0$

find $\frac{\text{d}y}{\text{d}x}$, giving your answer in its simplest form.

The curve $C$ has equation

$y={\text{e}}^{3x^2+5x-2}x\in \mathrm{ℝ}$

The point $P(-2,1)$ lies on $C$.

Find the equation of the tangent to $C$ at the point $P$, giving your answer in the form $ax+by+c=0$, where $a$, $b$ and $c$ are integers to be found.

Given that

$y=(4 \text{cos}x-3 \text{sin}x) {\text{e}}^{3x-5}x\in \mathrm{ℝ}$

find $\frac{\text{d}y}{\text{d}x}$, giving your answer in its simplest form.

Given that

$y=(x^3-4x^2+7)\ln xx>0$

find $\frac{\text{d}y}{\text{d}x}$, giving your answer in its simplest form.

Given that

$y=\frac{5x^2-10}{2x+1},x\ne 0.5$

Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{10x^2+10x+20}{{\left(2x+1\right)}^2}$.

Hence show that $y$ is an increasing function for all defined values of $x$.

Given that

$y=\frac{5x^7}{\text{sin}2x}0<x<\frac{\pi }{2}$

find $\frac{\text{d}y}{\text{d}x}$, giving your answer in its simplest form.

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

Given that $\theta$ is measured in radians, prove, from first principles, that

$\frac{\text{d}}{\text{d}\theta }\left(\text{cos}\theta \right)=-\text{sin}\theta$

You may assume the formula for $\text{cos}(A\pm B)$ and that as $h\to 0$, $\frac{\text{sin}h}{h}\to 1$ and $\frac{\text{cos}h-1}{h}\to 0$.

The curve $C$ has equation

$y={\text{e}}^{-3x}+\ln xx>0$

Show that the equation of the tangent to $C$ at the point where $x=1$ is

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

The curve $C$ has equation

$y=5 \text{cos}\left({\text{e}}^x-\frac{\pi }{2}\right)x\in \mathrm{ℝ}$

Find the gradient of the normal to $C$ at the point where $x=0$, giving your answer to 3 decimal places.

Given that

$y=(2 \text{sin}3x-\text{cos}3x) {\text{e}}^{6-x}x\in \mathrm{ℝ}$

find $\frac{\text{d}y}{\text{d}x}$, giving your answer in its simplest form.

Given that

$y=(x^2-x{)}^2\ln 5xx>0$

find $\frac{\text{d}y}{\text{d}x}$, giving your answer in its simplest form.

Given that

$x=\text{sec}7y0<y<\frac{\pi }{14}$

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

Hence show that

$\frac{\text{d}y}{\text{d}x}=\frac{1}{7x\sqrt{x^2-1}}$

Figure 2 shows a sketch of part of the curve with equation $y=\text{f}\left(x\right)$, where

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

The point $A$, shown in Figure 2, is a maximum turning point on the curve.

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

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

The curve $C$ has equation

$y=3^x+2^{-x}x\in \mathrm{ℝ}$

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

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

The function $\text{f}$ is defined by

$\text{f}\left(x\right)=\text{sin}\left(\text{cos}\left(\ln \frac{1}{x}\right)\right)x>0$

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

Given that $x$ is measured in radians, prove, from first principles, that the derivative of $\text{tan}3x$ is $3 {\text{sec}}^{2}3x$.

You may assume the formulae for $\text{sin}(A\pm B)$, $\text{cos}(A\pm B)$ and that as $h\to 0$, $\frac{\text{sin}h}{h}\to 1$ and $\frac{\text{cos}h-1}{h}\to 0$.

The curve $C$ has equation

$y=4^{-x^4}x\in \mathrm{ℝ}$

Show that

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

Hence find the equation of the tangent to $C$ 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.

Given that

$y=(5+{\text{sin}}^{2}3x) {\text{e }}^{x^2-3x+2}x\in \mathrm{ℝ}$

find $\frac{\text{d}y}{\text{d}x}$, giving your answer in its simplest form.

Given that

$y=3^{\sqrt{x}}\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)x>0$

find $\frac{\text{d}y}{\text{d}x}$, giving your answer in its simplest form.

Figure 1 shows a sketch of part of the curve with equation $y=\text{f}\left(x\right)$, where

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

The points $A$ and $B$, shown in Figure 1, are the maximum and minimum turning points on the curve respectively. The curve crosses the $x$-axis at the origin and at the point $\left(\frac{2\pi }{3}, 0\right)$.

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

The curve $C$ has equation

$y=\text{arctan}xx\in \mathrm{ℝ}$

The point $A$ lies on $C$.

The tangent to $C$ at the point $A$ passes through the point $\left(0, \frac{1}{2}\right)$.

Show that the $x$-coordinate of $A$ satisfies the equation

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

A sequence of functions $u_1,u_2,u_3,\dots$ is defined by the recurrence relation

$u_{k+1}\left(x\right)=\frac{\text{d}}{\text{d}x}\left(u_k\left(x\right)\right)k⩾1$

where

$u_1\left(x\right)=\text{sin}\left(x\sqrt{2}\right)$

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

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

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