Differentiation (Cambridge International AS Mathematics): Exam Questions

Differentiate

(i) $5 x$,

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

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

By first defining an appropriate function $u$, where $u$ is a function of $x$, show that the function $y$ defined by

$y = ( x - 7 )^{3}$

may be written in the form

$y = u^{n}$

where $n$ is a positive integer.

Using your result from part (a) along with the chain rule

$\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y}{\text{d}u} \times \frac{\text{d}u}{\text{d}x}$

find $\frac{\text{d}y}{\text{d}x}$ for the function $y$ defined in part (a). Be sure to give your answer entirely in terms of $x$.

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

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

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

Given that $y = ( x - 3 )^{4}$,

By first defining an appropriate function $u$ and writing $y$ in the form $y = u^{n}$ where $n$ is a positive integer, use the chain rule to find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$.

(i) By solving the equation $\frac{\text{d}y}{\text{d}x} = 0$, find the coordinates of any stationary points.

(ii) By considering the value of $\frac{\text{d}y}{\text{d}x}$ on either side of any stationary points, determine the nature of those stationary points.

Sketch the curve of $y = ( x - 3 )^{4}$.

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

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

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

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

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

Find $\frac{\text{d}y}{\text{d}x}$ when y = \left(\sqrt{x}\right)^{3} + \dfrac{2}{\sqrt{x}}.

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

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

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

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

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

For each of the following, find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$:

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

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

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

By rearranging the equation to make $y$ the subject, find $\frac{\text{d}y}{\text{d}x}$.

For each of the following, find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$:

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

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

For each of the following, find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$:

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

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

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

For each of the following, find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$:

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

y = x^{3}\left(\dfrac{1}{x^{3}} - \dfrac{2}{x^{2}} + \dfrac{3}{x}\right)

By first defining an appropriate function $u$, where $u = u ( x )$ is a function of $x$, show that the function $y$ defined by

$y = ( 3 x + 11 )^{5}$

may be written as a function of $u$.

Using your result from part (a) along with the chain rule

$\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y}{\text{d}u} \times \frac{\text{d}u}{\text{d}x}$

find $\frac{\text{d}y}{\text{d}x}$ for the function $y$ defined in part (a). Be sure to give your answer entirely in terms of $x$.

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

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

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

A curve has the equation $y = 3 x - 4 x^{-2}$, $x \neq 0$.

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

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

Given that $y = ( 2 x + 1 )^{4}$

use the chain rule to find $\frac{\text{d}y}{\text{d}x}$

Find the coordinates of any stationary points and determine their nature

Sketch the curve.

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

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

Given that the equation $\text{f} ' ( x ) = 0$ 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 $\frac{\text{d}y}{\text{d}x}$.

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

The curve with equation $y = a x^{2} + b x + c$ has a gradient of $- 7$ at the point $( - 1 , 13 )$, and a gradient of $- 3$ at the point $( 1 , 3 )$.

By considering $\frac{\text{d}y}{\text{d}x}$ show that $2 a + b = - 3$ and $- 2 a + b = - 7$.

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

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

Given that y = \dfrac{1}{\sqrt{x}}\left(1 + \dfrac{1}{x}\right), $x > 0$, find $\frac{\text{d}y}{\text{d}x}$.

For each of the following, find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$:

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

y = \dfrac{1}{x^{5}}\left(x^{2} + \sqrt{x} - 1\right)

For each of the following, use the chain rule to find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$:

y = \left(\dfrac{1}{x} + 2x\right)^{4}

y = \dfrac{1}{\left(x^{3} - 1\right)^{2}}

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

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

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

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

Given that $y = ( x^{2} - 2 x - 3 )^{4}$

use the chain rule to find $\frac{\text{d}y}{\text{d}x}$

Find the coordinates of any stationary points and determine their nature.

Sketch the curve.

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

The curve with equation $y = a x^{2} + b x + c$ 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 $c$.

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 $0$.

Given that y = \left(\dfrac{1}{x} - \dfrac{1}{x\sqrt{x}}\right)^{2}, $x > 0$, find $\frac{\text{d}y}{\text{d}x}$.

For each of the following, find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$:

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

$y = ( \sqrt{x} + 3 - \frac{4}{ \sqrt{x} } )^{2}$

For each of the following, use the chain rule to find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$:

y = \left(\dfrac{1}{\sqrt{x}} + 3x^{2}\right)^{3}

$y = \sqrt{\frac{1}{\sqrt{x}+\frac{1}{\sqrt{x}}}}$

The function $\text{f}$ is defined by $\text{f} ( x ) = 2 x^{3} + p x^{2} + 3 x - 16$. Determine the range of values for $p$ for which the equation $\text{f} ' ( x ) = 0$ has at least one real solution.

Given that y = \left(\dfrac{1}{x} + x\right)^{4}, $x \neq 0$

use the chain rule to find $\frac{\text{d}y}{\text{d}x}$

Find the coordinates of any stationary points and determine their nature.

Sketch the curve.

The function $\text{f}$ is defined by $\text{f} ( x ) = x^{n} - x$, $n \in ℕ$, $n \geq 2$. Determine the relationship between the value of $n$ and the number of real solutions to the equation $\text{f} ' ( x ) = 0$.

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

The curve with equation $y = a x^{2} + b x + c$ 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 $c$.