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

Given that

$y=\frac{3\sin \theta }{2\sin \theta +2\cos \theta }-\frac{\pi }{4}<\theta <\frac{3\pi }{4}$

show that

$\frac{\mathrm{d}y}{\mathrm{d}\theta }=\frac{A}{1+\sin 2\theta }-\frac{\pi }{4}<\theta <\frac{3\pi }{4}$

where $A$ is a rational constant to be found.

Figure 1 shows a sketch of the curve $C$ with equation

$y=\frac{4x^2+x}{2\sqrt{x}}-4\ln xx>0$

Show that

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{12x^2+x-16\sqrt{x}}{4x\sqrt{x}}$

The point $P$, shown in Figure 1, is the minimum turning point on $C$.

Show that the $x$ coordinate of $P$ is a solution of

$x={\left(\frac{4}{3}-\frac{\sqrt{x}}{12}\right)}^{\frac{2}{3}}$

A car stops at two sets of traffic lights.

Figure 2 shows a graph of the speed of the car, $v{\mathrm{ms}}^{-1}$, as it travels between the two sets of traffic lights.

The car takes $T$ seconds to travel between the two sets of traffic lights.

The speed of the car is modelled by the equation

$v=(10-0.4t)\ln (t+1)0\le t\le T$

where $t$ seconds is the time after the car leaves the first set of traffic lights.

According to the model, find the value of $T$

Show that the maximum speed of the car occurs when

$t=\frac{26}{1+\ln \left(t+1\right)}-1$

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

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

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

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

$\mathrm{f}\left(x\right)=\frac{{\mathrm{e}}^{3x}}{4x^2+k}$

where $k$ is a positive constant.

Show that

${\mathrm{f}}^{\text{'}}\left(x\right)=\left(12x^2-8x+3k\right)\mathrm{g}\left(x\right)$

where $\mathrm{g}\left(x\right)$ is a function to be found.

Given that the curve with equation $y=\mathrm{f}\left(x\right)$ has at least one stationary point, find the range of possible values of $k$.

$y=\sin x$

where $x$ is measured in radians.

Use differentiation from first principles to show that

$\frac{\mathrm{d}y}{\mathrm{d}x}=\cos x$

You may

• use without proof the formula for $\sin \left(A\pm B\right)$

• assume that as $h\to 0$, $\frac{\sin h}{h}\to 1$ and $\frac{\cos h-1}{h}\to 0$

The function $\mathrm{g}$ is defined by

$\mathrm{g}\left(x\right)=\frac{3\ln \left(x\right)-7}{\ln \left(x\right)-2}x>0x\ne k$

where $k$ is a constant.

Deduce the value of $k$.

Prove that

${\mathrm{g}}^{\text{'}}\left(x\right)>0$

for all values of $x$ in the domain of $\mathrm{g}$.

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

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

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

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 $\left(\frac{\pi }{3}, \frac{2\sqrt{3}}{3}\right)$.

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

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

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.

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

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

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

$\mathrm{f}\left(x\right)=4\left(x^2-2\right){\mathrm{e}}^{-2x}x\in \mathrm{ℝ}$

Show that $\mathrm{f}\text{'}\left(x\right)=8\left(2+x-x^2\right){\mathrm{e}}^{-2x}$.

Hence find, in simplest form, the exact coordinates of the stationary points of $C$.

The function $\mathrm{g}$ and the function $\mathrm{h}$ are defined by

$\begin{array}{l}\mathrm{g}\left(x\right)=2\mathrm{f}\left(x\right)x\in \mathrm{ℝ}\\ \mathrm{h}\left(x\right)=2\mathrm{f}\left(x\right)-3x\ge 0\end{array}$

Find

(i) the range of $\mathrm{g}$

(ii) the range of $\mathrm{h}$.

$\frac{1+11x-6x^2}{\left(x-3\right)\left(1-2x\right)}\equiv A+\frac{B}{\left(x-3\right)}+\frac{C}{\left(1-2x\right)}$

Find the values of the constants $A$, $B$ and $C$.

$\mathrm{f}\left(x\right)=\frac{1+11x-6x^2}{\left(x-3\right)\left(1-2x\right)}x>3$

Prove that $\mathrm{f}\left(x\right)$ is a decreasing function.

Given that

$y=\frac{x-4}{2+\sqrt{x}}x>0$

show that

$\frac{dy}{dx}=\frac{1}{A\sqrt{x}}x>0$

where $A$ is a constant to be found.

A curve has equation $y=\mathrm{f}\left(x\right)$, where

$\mathrm{f}\left(x\right)=\frac{7x{\mathrm{e}}^x}{\sqrt{{\mathrm{e}}^{3x}-2}}x>\ln \sqrt[3]{2}$

Show that

${\mathrm{f}}^{\text{'}}\left(x\right)=\frac{7{\mathrm{e}}^x\left({\mathrm{e}}^{3x}\left(2-x\right)+Ax+B\right)}{2{\left({\mathrm{e}}^{3x}-2\right)}^{{\displaystyle \frac{3}{2}}}}$

where $A$ and $B$ are constants to be found.

Hence show that the $x$ coordinates of the turning points of the curve are solutions of the equation

$x=\frac{2{\begin{array}{rcl}& & \mathrm{e}\end{array}}^{3x}-4}{{\begin{array}{rcl}& & \mathrm{e}\end{array}}^{3x}+4}$

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

$\mathrm{f}\left(x\right)=10{\mathrm{e}}^{-0.25x}\sin x$

Show that the $x$-coordinates of the turning points of the curve with equation $y=\mathrm{f}\left(x\right)$ satisfy the equation $\tan x=4$

Figure 3 shows a sketch of part of the curve with equation $y=\mathrm{f}\left(x\right)$.

Sketch the graph of $H$ against $t$ where

$H\left(t\right)=\left|10{\mathrm{e}}^{-0.25t}\sin t\right|$

showing the long-term behaviour of this curve.

The function $H\left(t\right)$ is used to model the height, in metres, of a ball above the ground $t$ seconds after it has been kicked.

Using this model, find the maximum height of the ball above the ground between the first and second bounce.

The curve $C$, in the standard Cartesian plane, is defined by the equation

$\begin{array}{ccc}x=4\sin 2y& & -\frac{\pi }{4}<y<\frac{\pi }{4}\end{array}$

The curve passes through the origin $O$

Find the value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ at the origin.

(i) Use the small angle approximation for $\sin 2y$ to find an equation linking $x$ and $y$ for points close to the origin.

(ii) Explain the relationship between the answers to (a) and (b)(i).

Show that, for all points $\left(x,y\right)$ lying on $C$,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{a\sqrt{b-x^2}}$

where $a$ and $b$ are constants to be found.

A scientist is studying a population of mice on an island.

The number of mice, $N$, in the population, $t$ months after the start of the study, is modelled by the equation

$N=\frac{900}{3+7{\mathrm{e}}^{-0.25t}},t\in \mathrm{ℝ},t\ge 0$

Find the number of mice in the population at the start of the study.

Show that the rate of growth $\frac{dN}{dt}$ is given by $\frac{dN}{dt}=\frac{N\left(300-N\right)}{1200}$

The rate of growth is a maximum after $T$ months.

Find, according to the model, the value of $T$.

According to the model, the maximum number of mice on the island is $P$.

State the value of $P$.

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

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