Differentiation (Cambridge International IGCSE Further Maths): Exam Questions

Given that $y=\mathrm{tan}x$, use calculus to find the approximate change in $y$ as $x$ increases from $-\frac{π}{4}$ to $h-\frac{π}{4}$, where $h$ is small.

Find the $x$-coordinate of the stationary point on the curve $y = ( 2 - \sqrt{3} ) x^{2} + x - 1$, giving your answer in the form $a+b\sqrt{3}$,where $a$ and $b$ are rational numbers.

The radius, $r\mathrm{cm}$, of a circle is increasing at the rate of $5\mathrm{cms}^{--1}$. Find, in terms of $π$, the rate at which the area of the circle is increasing when $r=3$.

The volume, $V$, of a sphere of radius $r$ is given by $V=\frac{4}{3}\mathrm{πr}^{3}$. The radius, $r\mathrm{cm}$, of a sphere is increasing at the rate of $0.5\mathrm{cms}^{-1}$. Find, in terms of $π$, the rate of change of the volume of the sphere when $r=0.25$.

Given that $y = ( x^{2} - 1 ) \sqrt{5 x + 2}$, show that $\frac{dy}{dx}=\frac{Ax^{2}+Bx+C}{2\sqrt{5x+2}}$, where $A,B$ and $C$ are integers.

Extract:

Given that y = ( x ^2 - 1 ) sqrt( 5 x + 2 ), show that ( d y)/( d x) = (A x ^2 + B x + C)/(2 sqrt( 5 x + 2 )), where A , B and C are integers.

$The \text{derivative} is: \frac{ \mathrm{d} y}{ \mathrm{d} x} = \frac{25 x ^{2} + 8 x - 5}{2 \sqrt{5 x + 2}}$

Find the coordinates of the stationary point of the curve $y = ( x ^{2} - 1 ) \sqrt{5 x + 2} \text{for} x > 0$. Give each coordinate correct to 2 significant figures.

Extract:

$Given \text{that} y = ( x ^{2} - 1 ) \sqrt{5 x + 2} , \text{from part} (a): \frac{ \mathrm{d} y}{ \mathrm{d} x} = \frac{25 x ^{2} + 8 x - 5}{2 \sqrt{5 x + 2}}$

From part (b): The stationary point of the curve y = ( x ^2 - 1 ) sqrt( 5 x + 2 ) for x greater than 0 occurs at x = 0.315 and y = - 1.70 (coordinates to 2 significant figures).

Determine the nature of this stationary point.

Variables $x$ and $y$ are such that $y=\mathrm{sin}x+e^{-x}$. Use differentiation to find the approximate change in $y$ as $x$increases from $\frac{π}{4}\mathrm{to}\frac{π}{4}+h$, where h is small.

The equation of a curve is $y=x\sqrt{16-x^{2}}\mathrm{for}0\leq x\leq4$.

Find the exact coordinates of the stationary point of the curve.

Find $\frac{ \mathrm{d} }{ \mathrm{d} x} ( 16 - x^{2} )^{ \frac{3}{2} }$and hence evaluate the area enclosed by the curve $y=x\sqrt{16-x^{2}}$ and the lines $y=0,x=1\mathrm{and}x=3$.

A curve has equation $y=(2x-1)\sqrt{4x+3}$.

Show that $\frac{ \mathrm{d} y}{ \mathrm{d} x} = \frac{4 ( A x + B )}{ \sqrt{4 x + 3} }$, where $A$ and $B$ are constants.

Extract:

A curve has equation y = (2x-1)√(4x+3).

Show that dy/dx = 4(Ax+B)/√(4x+3), where A and B are constants.

From the mark scheme: dy/dx = 4(3x+1)/√(4x+3)

Hence write down the $x$-coordinate of the stationary point of the curve.

Extract:

$A \text{curve has equation} y = ( 2 x - 1 ) \sqrt{4 x + 3} .$

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = \frac{4 ( 3 x + 1 ) }{\sqrt{4 x + 3}}$

The x-coordinate of the stationary point of the curve is x equals negative 1 third.

Determine the nature of this stationary point.

Find the equation of the tangent to the curve $y=x^{3}-6x^{2}+3x+10$ at the point where $x=1$.

Extract:

Find the equation of the tangent to the curve y = x³ - 6x² + 3x + 10 at the point where x = 1.

The equation of the tangent is y = -6x + 14

Find the coordinates of the point where this tangent meets the curve again.

It is given that $y=\mathrm{ln}(1+\mathrm{sin}x)$ for $0<x<π$ .

Find $\frac{dy}{dx}$

Extract:

It is given that y = ln(1 + sin x) for 0 < x < π.

Find dy/dx

To differentiate y = ln(1 + sin x), we can use the fact that the differential of ln x is 1/x and chain rule.

dy/dx = 1/(1 + sin x) × cos x

dy/dx = (cos x)/(1 + sin x)

Find the value of $\frac{dy}{dx}$ when $x=\frac{π}{6}$, giving your answer in the form $\frac{1}{\sqrt{a}}$, where $a$ is an integer.

Extract:

It is given that y = ln(1 + sin x) for 0 < x < π.

Find dy/dx

dy/dx = cos x / (1 + sin x)

Find the values of $x$ for which $\frac{dy}{dx}=\mathrm{tan}x$.

A curve has equation $y=x\mathrm{cos}x$.

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

Extract:

A curve has equation y = x cos x.

dy/dx = -x sin x + cos x

Find the equation of the normal to the curve at the point where $x=π$, giving your answer in the form $y=mx+c$.

Find the equation of the tangent to the curve $y = \frac{\ln ( 3 x^{2} - 1 )}{x + 2}$ at the point where $x=1$. Give your answer in the form$y=mx+c$, where $m$ and $c$ are constants correct to 3 decimal places.

A curve has equation $y=\mathrm{ln}(5--3x)$ where $x<\frac{5}{3}$. The normal to the curve at the point where $x=--5$ , cuts the $x$-axis, at the point $P$. Find the equation of the normal and the $x$-coordinate of $P$.

Variables $x$ and $y$ are such that $y=e^{\frac{x}{2}}+x\mathrm{cos}2x$ , where $x$ is in radians. Use differentiation to find the approximate change in $y$ as $x$ increases from 1 to $1+h$, where $h$ is small.

The tangent to the curve $y = \ln ( 3 x^{2} - 4 ) - \frac{x^{3}}{6}$, at the point where $x=2$, meets the $y$-axis at the point $P$. Find the exact coordinates of $P$.

A container is a circular cylinder, open at one end, with a base radius of $r$ cm and a height of $h$ cm. The volume of the container is 1000 cm³. Given that $r$ and $h$ can vary and that the total outer surface area of the container has a minimum value, find this value.

Find the $x$-coordinates of the stationary points of the curve $y=e^{3x}(2x+3)^{6}$.

A curve has equation $y=f(x)$ and has exactly two stationary points. Given that $f^{"}(x)=4x-7$, $f'(0.5)=0\mathrm{and}f'(3)=0$, use the second derivative test to determine the nature of each of the stationary points of this curve.

In this question all lengths are in centimetres.

The diagram shows a solid cuboid with height $h$ and a rectangular base measuring $4x$ by $x$. The volume of the cuboid is $40\mathrm{cm}^{3}$. Given that $x$ and $h$ can vary and that the surface area of the cuboid has a minimum value, find this value.

Find the equation of the tangent to the curve $2y=\mathrm{tan}2x+7$ at the point where $x=\frac{π}{8}$. Give your answer in the form $ax-y=\frac{π}{b}+c$, where $a,b$ and $c$ are integers.

Extract:

Find the equation of the tangent to the curve 2y = tan 2x + 7 at the point where x = π/8.

The equation of the tangent is 2x - y = π/4 - 4

This tangent intersects the $x$-axis at $P$ and the $y$-axis at $Q$. Find the length of $PQ$.

$y=x\sqrt{x+2}$ Given that , show that $\frac{dy}{dx}=\frac{Ax+B}{2\sqrt{x+2}}$, where $A$ and $B$ are constants.

Extract:

y = x√(x+2)

From part (a): dy/dx = (3x+4)/(2√(x+2))

Find the exact coordinates of the stationary point of the curve $y=x\sqrt{x+2}$.

Extract:

y = x√(x+2)

dy/dx = (3x + 4)/(2√(x+2))

Stationary point occurs at x = -4/3

Determine the nature of this stationary point.

Differentiate $y=\mathrm{tan}(x+4)-3\mathrm{sin}x$ with respect to $x$.

Variables $x$ and $y$ are such that $y = \frac{\ln ( 2 x + 5 )}{2 e^{3 x} }$. Use differentiation to find the approximate change in $y$ as $x$ increases from $1\mathrm{to}1+h$, where $h$ is small.

It is given that $y=\frac{\mathrm{tan}3x}{\mathrm{sin}x}$.

Find the exact value of $\frac{dy}{dx}$ when $x=\frac{π}{3}$.

Extract:

$It is \text{given that} y = \frac{ \text{tan} 3 x}{ \text{sin} x}.$

Find the exact value of ( d y)/( d x) when x = pi / 3.

At x = π/3, dy/dx = 2√3

Hence find the approximate change in $y$ as $x$ increases from $\frac{π}{3}\mathrm{to}\frac{π}{3}+h,$ where $h$ is small.

Extract:

$It is \text{given that} y = \frac{ \text{tan} 3 x}{ \text{sin} x}.$

Find the exact value of ( d y)/( d x) when x = pi / 3.

Given that $x$ is increasing at the rate of 3 units per second, find the corresponding rate of change in $y$ when $x=\frac{π}{3}$ , giving your answer in its simplest surd form.

It is given that $y=\mathrm{ln}(\mathrm{sin}x+3\mathrm{cos}x)\mathrm{for}0<x<\frac{π}{2}$.

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

Extract:

y = ln(sin x + 3 cos x) for 0 < x < π/2

dy/dx = (cos x - 3 sin x)/(sin x + 3 cos x)

Find the value of $x$ for which $\frac{dy}{dx}=-\frac{1}{2}$.

Given that $y=\frac{e^{2x-3}}{x^{2}+1}$, find $\frac{dy}{dx}$.

Extract:

$Given \text{that} y = \frac{ \mathrm{e} ^{2} x - 3}{x ^{2} + 1}, \text{find} \frac{ \mathrm{d} y}{ \mathrm{d} x}.$

Differentiate the equation using the quotient rule and chain rule.

$u = \mathrm{e} ^{2} x - 3 v = x ^{2} + 1$

$\frac{ \mathrm{d} u}{ \mathrm{d} x} = 2 \mathrm{e} ^{2} x - 3 \frac{ \mathrm{d} v}{ \mathrm{d} x} = 2 x$

Substitute into the quotient rule, ( d y)/( d x) = (( v ) ( ( d u)/( d x) ) - ( u ) ( ( d v)/( d x) ))/(v ^2).

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = \frac{ ( x ^{2} + 1 ) ( 2 \mathrm{e} ^{2} x - 3 ) - ( 2 x ) ( \mathrm{e} ^{2} x - 3 ) }{ ( x ^{2} + 1 ) ^{2} }$

Hence, given that $y$ is increasing at the rate of 2 units per second, find the exact rate of change of $x$ when $x=2$.

A sphere of radius $r$ cm and volume $V$ cm³ is increasing in size with time $t$ seconds. The volume increases at a constant rate of 24 cm³ s^-1.

Find the exact rate at which the radius is increasing when the sphere reaches a volume of $\frac{32π}{3}$ cm³.

A sector from a circle of radius $r$ has an internal angle of $θ$ radians, as shown below.

The perimeter of the sector is 4 units and the area of the sector is $A$ square units.

Show that $A = \frac{8 \\theta }{( 2 + \\theta )^{2}}$.

Show that $\frac{ \mathrm{d} A}{ \mathrm{d} \\theta } = \frac{p ( q - \\theta )}{( 2 + \\theta )^{3}}$ where $p$ and $q$ are constants to be found.

Find the maximum area of the sector. You must show that this area is a maximum.

A curve is given by $y = \ln ( 1 + x^{4} )$.

Use calculus to find the approximate change in $y$ as $x$ increases from $-1$ to $-1+k$ where $k$ is small.

In this question, all lengths are in centimetres.

The diagram shows a cone of base radius $x$, height $y$ and sloping edge $\sqrt{x^{2}+y^{2}}$. The volume of the cone is $10π\mathrm{cm}^{3}$.

Show that the curved surface area, $S$, of the cone is given by $S=\frac{π\sqrt{x^{6}+900}}{x}$.

Extract:

In this question, all lengths are in centimetres.

The diagram shows a cone of base radius x, height y and sloping edge √(x² + y²). The volume of the cone is 10π cm³.

Show that the curved surface area, S, of the cone is given by S = (π√(x^6 + 900))/x.

Given that $x$ can vary and that $S$ has a minimum value, find the value of $x$ for which $S$ is a minimum.

Variables $x$ and $y$ are such that $y=\frac{e^{3x}\mathrm{sin}x}{x^{2}}$ .Use differentiation to find the approximate change in $y$as $x$ increases from $0.5\mathrm{to}0.5+h$, where $h$is small.

In this question all lengths are in centimetres. The volume, $V$, of a cone of height $h$ and base radius $r$ is given by $V=\frac{1}{3}πr^{2}h$

The diagram shows a large hollow cone from which a smaller cone of height 180 and base radius 90 has been removed. The remainder has been fitted with a circular base of radius 90 to form a container for water. The depth of water in the container is w and the surface of the water is a circle of radius $R$.

Find an expression for $R$ in terms of $w$ and show that the volume $V$ of the water in the container is given by V = pi / 12 ( w + 180 ) ^3 - 486000 pi.

Extract:

The diagram shows a large hollow cone from which a smaller cone of height 180 and base radius 90 has been removed. The remainder has been fitted with a circular base of radius 90 to form a container for water. The depth of water in the container is w and the surface of the water is a circle of radius R.

Find an expression for R in terms of w and show that the volume V of the water in the container is given by V = (π/12)(w + 180)³ - 486000π.

Water is poured into the container at a rate of $10000\mathrm{cm}^{3}s^{-1}$. Find the rate at which the depth of the water is increasing when $w=10$.

A curve has equation $y = \frac{\ln ( 3 x^{2} - 5 )}{2 x + 1} \text{for} 3 x^{2} > 5$

Find the equation of the normal to the curve at the point where $x=\sqrt{2}$.

Extract:

$A \text{curve has equation} y = \frac{ln ( 3 x ^{2} - 5 ) }{2 x + 1} \text{for} 3 x ^{2} > 5$

At the point where x space equals space square root of 2:

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = \frac{6 \sqrt{2} }{2 \sqrt{2} + 1}$

Find the approximate change in $y$ as $x$ increases from $\sqrt{2}\mathrm{to}\sqrt{2}+h$, where $h$ is small.

The rectangle $ABCDE$ represents a ploughed field where $AB=300m$ and $AE=400m$. Joseph needs to walk from $A$ to $D$ in the least possible time. He can walk at $0.9\mathrm{ms}^{-1}$ on the ploughed field and at $1.5\mathrm{ms}^{-1}$ on any part of the path $BCD$ along the edge of the field. He walks from $A$ to $C$ and then from $C$ to $D$. The distance $BC=xm$.

Find, in terms of $x$, the total time, $Ts$, Joseph takes for the journey.

Extract:

The rectangle ABCDE represents a ploughed field where AB = 300 m and AE = 400 m. Joseph needs to walk from A to D in the least possible time. He can walk at 0.9 ms⁻¹ on the ploughed field and at 1.5 ms⁻¹ on any part of the path BCD along the edge of the field. He walks from A to C and then from C to D. The distance BC = x m.

$T = \frac{10}{9} \sqrt{90 000 + x ^{2} } + \frac{2 ( 400 - x ) }{3}$

Given that $x$ can vary, find the value of $x$ for which $T$ is a minimum and hence find the minimum value of $T$.

A curve has the equation $y = \mathrm{e}^{( 4 x - x^{2} )}$.

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

Hence find $\int \frac{( 2 - x ) \mathrm{e}^{4 x} }{ \mathrm{e}^{x^{2}} } \mathrm{d} x$.

Show that the second derivative of the curve $y = \mathrm{e}^{( 4 x - x^{2} )}$ satisfies the relationship

$\frac{ \mathrm{d}^{2} y}{ \mathrm{d} x^{2}} = ( p + q x + r x^{2} ) y$

where $p$, $q$ and $r$ are constants to be found.