Differentiation (Edexcel IAL Maths): Exam Questions

Differentiate

(i) $5x,$

(ii) $2x^{3},$

(iii) $x^{\frac{1}{2}}.$

Write down the gradient of the line with equation $y=k$, where $k$ is a constant.

Find the gradient at the point where $x=8$ for the following functions

(i) $\mathrm{f} ( \mathrm{x} ) = 3 \mathrm{x}^{2} ,$

(ii) $\mathrm{f} ( \mathrm{x} ) = 4 \mathrm{x}^{3} - 2 \mathrm{x} ,$

(iii) $\mathrm{f} ( \mathrm{x} ) = 3 \mathrm{x}^{1 \frac{1}{3}} .$

(i) Expand $( x + 3 ) ( x - 2 )$.

(ii) Hence differentiate $( x + 3 ) ( x - 2 )$.

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

Find the $x$-coordinate of the point on the curve $y=5x^{2}-16x$ where the gradient is 4.

Find the coordinates of the points on the curve $y=2x^{3}-9x^{2}+12x$ where the gradient is 0.

Find $\frac{dy}{dx}$ when $y = ( \sqrt{x} )^{3} + \frac{2}{ \sqrt{x} }$.

The function $\mathrm{f} ( \mathrm{x} )$ is given by

$\mathrm{f} ( \mathrm{x} ) = \frac{2 \mathrm{x}^{1 \frac{1}{3}} + 3 \mathrm{x}^{ \frac{2}{3} } }{ \mathrm{x} }$.

Show that $\mathrm{f} ( \mathrm{x} )$ can be written in the form $\mathrm{f} ( x ) = a x^{b} + c x^{d}$, where a, b, c and d are constants to be found.

Find $\mathrm{f}^{' } ( \mathrm{x} ).$

Find an expression for $\frac{dy}{dx}$ when $y=3x^{2}-2x$.

Find the gradient of $y=3x^{2}-2x$ at the points where

(i) $x=3,$

(ii) $x=-2$.

(i) Find the gradient of the tangent at the point $( 2 , 3 )$ on the graph of $y=2x^{3}-3x^{2}-1$.

(ii) Hence find the equation of the tangent at the point $( 2 , 3 )$.

For the graph with equation $y=3x-\frac{1}{2}x^{2}$, find the gradient of the tangent at the point where $x=5$.

(i) Find the gradient of the normal at the point where $x=5$.

(ii) Hence find the equation of the normal at the point where $x=5$.

For each of the following, find $\frac{dy}{dx}$ in terms of x:

$y=4x^{2}-3x+19$

$y=x^{3}-5x^{2}+14x-1$

$y=4x^{\frac{3}{2}}-3x^{-1}$

Given that $y=\sqrt{x}+\frac{1}{\sqrt{x}},x>0,$ find $\frac{dy}{dx}$.

For each of the following, find $\frac{dy}{dx}$ in terms of x:

$y = ( 2 x + 3 ) ( 3 x - 1 )$

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

The function f is defined by $\mathrm{f} ( \mathrm{x} ) = 2 \mathrm{x}^{3 } - \mathrm{x}^{2} - 4 \mathrm{x} + 3$.

Find $\mathrm{f}^{' } ( x )$.

Solve the equation $\mathrm{f}^{' } ( x ) = 0$.

A curve has the equation $y=3x-4x^{-2},x\neq0$.

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

Find the coordinates of the point on the curve where the gradient is 2.

The function f is defined by $\mathrm{f} ( \mathrm{x} ) = \mathrm{x}^{3} - 6 \mathrm{x}^{2} - \text{cx} + 12$.

Find $\mathrm{f}^{' } ( \mathrm{x} ).$

Given that the equation $\mathrm{f}^{' } ( \mathrm{x} ) = 0$ has exactly one real solution, find the value of c.

A curve is described by the equation $\frac{y}{x-3}=x^{2}+1.$ * *

Make y the subject of the equation.

Hence find $\frac{dy}{dx}.$

Find the coordinates of the point on the curve where the gradient is $-2$.

The curve with equation $y=ax^{2}+bx+c$ has a gradient of $-7$ at the point $( - 1 , 13 ) ,$ and a gradient of $-3$ at the point $( 1 , 3 )$. By considering $\frac{dy}{dx}$ show that $2a+b=-3$ and $-2a+b=-7.$

Hence find the values of $a$ and $b$.

By considering a point that you know to be on the curve, find the value of $c$.

The curve $C$ has equation $y=2x^{3}-3x^{2}+4x-3.$

Show that the point $P ( 2 , 9 )$ lies on $C$ .

Show that the value of $\frac{dy}{dx}$ at $P$ is 16.

Find an equation of the tangent to $C$ at $P$.

The curve $C$ has equation $y=3x^{2}-6+\frac{4}{x}.$ The point $P ( 1 , 1 )$ lies on $C$.

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

Show that an equation of the normal to $C$ at point $P$ is $x+2y=3$.

This normal cuts the -axis at the point $Q$ .

Find the length of $PQ$, giving your answer as an exact value.

Given that $y=2x^{3}-8\sqrt{x}$, find

$\frac{dy}{dx}$

$\frac{d^{2}y}{dx^{2}}$

For each of the following, find $\frac{dy}{dx}$ in terms of $x$:

$y=-3x^{3}+5x^{2}-3x+\sqrt{13}$

$y=9x^{\frac{1}{3}}-6x^{-\frac{1}{3}}$

Given that $y = \frac{1}{ \sqrt{x} } ( 1 + \frac{1}{x} ) , x > 0 ,$find $\frac{dy}{dx}.$

For each of the following, find $\frac{dy}{dx}$ in terms of $x$:

$y = ( 2 x - 1 )^{2} ( x + 1 )$

$y=\frac{1}{x^{5}}(x^{2}+\sqrt{x}-1)$

The function $f$ is defined by $\mathrm{f} ( \mathrm{x} ) = \mathrm{x}^{3} - 4 \mathrm{x}^{2} + 6 \mathrm{x} - 9.$ Show that there are no solutions to the equation $\mathrm{f}^{' } ( \mathrm{x} ) = 0.$

A curve has the equation $y=\frac{3}{8}x^{\frac{4}{3}}-12x^{\frac{1}{3}}.$

Show that $\frac{d y}{d x} = a x^{- \frac{2}{3} } ( x + b ) ,$ where $a$ and $b$ are rational numbers to be found.

Hence find the coordinates of the point on the curve where the gradient is 0.

A curve has the equation $y=4x^{3}+bx^{2}+3x-17,$ where is a constant.* *Given that there is only one point on the curve where the gradient is zero, determine the possible values of $b$.

A curve is described by the equation $4y^{2}-3x^{5}=0,y>0$.

By rearranging the equation to make $y$ the subject, find $\frac{dy}{dx}$.

The curve with equation $y=ax^{2}+bx+c$ has a gradient of 8 at the point $( - 2 , 0 )$, and a gradient of $-10$ at the point $( 1 , - 3 )$. Find the values of $a,b$ and $c$.

The curve $C$ has equation $y=3x^{2}-6x+\sqrt{2x.}$ The point $P ( 2 , 2 )$ lies on $C$.

Find an equation of the tangent to $C$ at $P$.

The curve $C$ has equation $y=\frac{9}{\sqrt{3x}}-\frac{3}{x}.$ The point $P ( 3 , 2 )$ lies on $C$.

The normal to $C$ at $P$ intersects the $x$-axis at the point $Q$.

Find the coordinates of $Q$.

Given that $y = \frac{4}{x} - \sqrt[3]{\frac{27}{x}}$, find

$\frac{dy}{dx}$

$\frac{d^{2}y}{dx^{2}}$

For each of the following, find $\frac{dy}{dx}$ in terms of x :

$y=-\frac{5}{4}x^{3}+\frac{3}{5}x^{2}-x\sqrt{2}+π$

$y=\frac{3}{2}x^{\frac{4}{5}}-\frac{10}{3}x^{-\frac{4}{5}}$

Given that $y = ( \frac{1}{x} - \frac{1}{x \sqrt{x} } )^{2}$, $x>0,$ find $\frac{dy}{dx}.$

For each of the following, find $\frac{dy}{dx}$in terms of $x$:

$y=\frac{2x^{3}-5x^{2}-3x}{2x+1}$

A curve has the equation $y=x\sqrt{x}+\frac{48}{\sqrt{x}},x>0$. Find the coordinates of the point on the curve where the gradient is 0.

The function f is defined by $\mathrm{f} ( \mathrm{x} ) = \mathrm{x}^{ \mathrm{n} } - \mathrm{x} ,$ $n\inℕ,n\geq2.$ Determine the relationship between the value of $n$ and the number of real solutions to the equation $\mathrm{f}^{' } ( \mathrm{x} ) = 0$.

A curve is described by the equation $\frac{\sqrt{y}}{-1+\sqrt{x}}=\frac{1}{x},x>1.$ Find$\frac{dy}{dx}.$

The curve with equation $y=ax^{2}+bx+c$ passes through the point $( - 1 , 4 )$. At the point $( 2 , 7 )$ the gradient of the curve is 7. Find the values of $a,b$ and $c$.

A curve has equation $y=5-(x-3)^{2}$.

A is the point on the curve with $x$ coordinate 0, and $B$ is the point on the curve with $x$ coordinate 6.

$C$ is the point of intersection of the tangents to the curve at $A$ and $B$.

Find the coordinates of point $C$.

Calculate the area of triangle $ABC$.

A curve is described by the equation $y = \mathrm{f} ( \mathrm{x} )$, where

$\mathrm{f} ( \mathrm{x} ) = \frac{1}{ \sqrt{x} } , \mathrm{x} > 0$

$P$ is the point on the curve such that the normal to the curve at $P$ also passes through the origin.

Find the coordinates of point $P$. Give your answer in the form $( 2^{a} , 2^{b} )$, where a and b are rational numbers to be found.

Write down the equation of the normal to the curve at $P$.

Show that an equation of the tangent to the curve at $P$ is

$( 2^{1 \frac{1}{3}} ) x + ( 2^{ \frac{5}{6} } ) y = 3$