Differentiation (OCR AS Maths A: Pure): Exam Questions

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

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

Show that $\mathrm{f}\left(x\right)$ can be written in the form

$\mathrm{f}\left(x\right)=a{x}^b+c{x}^d$

where $a$, $b$, $c$ and $d$ are constants to be found.

Hence find an expression for $\mathrm{f}\text{'}\left(x\right)$.

Find the coordinates of any points on the curve

$y=2x^3-9x^2+12x$

at which the gradient is zero.

Find the $x$-coordinate of the point on the curve with equation

 $y=5x^2-16x$

at which the gradient is 4.

A student uses differentiation from first principles to show that the derivative of $7x^2$ is $14x$.

• Their working is shown below.

• There is an error in their argument in step 6.

Write out the correct argument for step 6.

STEP 1

$\mathrm{f}\left(x\right)=7x^2$

STEP 2

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

STEP 3

$\mathrm{f}\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{7{\left(x+h\right)}^2-7x^2}{h}$

STEP 4

$\mathrm{f}\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{7x^2+14hx+7h^2-7x^2}{h}$

STEP 5

$\mathrm{f}\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{h(14x+7h)}{h}$

STEP 6

$\mathrm{f}\text{'}\left(x\right)=14x+7h$

When  $h=0$,  $14x+7h=14x$

$\therefore \mathrm{f}\text{'}\left(x\right)=14x$

Given that

$\mathrm{f}\left(x\right)=4x$

Use differentiation from first principles to show that

$\mathrm{f}\text{'}\left(x\right)=4$

Given that

$\mathrm{f}\left(x\right)=-3x$

Use differentiation from first principles to show that

$\mathrm{f}\text{'}\left(x\right)=-3$

Given that

$y={\left(\sqrt{x}\right)}^3+\frac{2}{\sqrt{x}}$

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

Given that

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

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

A curve has equation

$y=\left(2x+3\right)\left(3x-1\right)$

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ writing your answer in simplest form.

A curve has equation

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

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ writing your answer in simplest form.

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

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

Find $\mathrm{f}\text{'}\left(x\right)$ writing your answer in simplest form.

Hence solve the equation

$\mathrm{f}\text{'}\left(x\right)=0$

A curve has the equation

$y=3x-{4x}^{-2}x\ne 0$

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

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

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

$\mathrm{f}\left(x\right)=x^3-6x^2-cx+12$

where $c$ is a constant.

Find an expression for $\mathrm{f}\text{'}\left(x\right)$ in terms of $x$ and $c$.

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

A curve has equation

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

Write the equation of the curve in the form

$y=a{x}^3+b{x}^2+cx+d$

where $a$, $b$, $c$ and $d$ are constants to be found.

Hence find $\frac{dy}{dx}$ writing your answer in simplest form.

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

A curve has the equation

$y=a{x}^2+bx+c$

where $a$, $b$ and $c$ are non-zero constants.

• The gradient at the point $\left(-1,13\right)$ is $-7$

• The gradient at the point $\left(1,3\right)$ is $-3$

Show that

$-2a+b=-7$

and

$2a+b=-3$

Hence find $a$ and $b$.

By considering the point $\left(-1,13\right)$, find $c$.

A curve has equation

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

Find $\frac{dy}{dx}$ writing your answer in simplest form.

A curve has equation

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

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

Given that

$\mathrm{f}\left(x\right)=a{x}^2$

Use differentiation from first principles to show that

$\mathrm{f}\text{'}\left(x\right)=2ax$

where $a$ is a constant.

Given that

$\mathrm{f}\left(x\right)=2x^3$

Use differentiation from first principles to show that

$\mathrm{f}\text{'}\left(x\right)=6x^2$

A curve has equation

$y=\frac{1}{\sqrt{x}}\left(1+\frac{1}{x}\right)$

Show that

$\frac{dy}{dx}=a{x}^b+c{x}^d$

where $a$, $b$, $c$ and $d$ are constants to be found.

A curve has equation

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

Find $\frac{dy}{dx}$ writing your answer in simplest form.

A curve has equation

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

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

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

$\mathrm{f}\left(x\right)=x^3-4x^2+6x-9$

Show that there are no solutions to the equation

$\mathrm{f}\text{'}\left(x\right)=0$

A curve has the equation

$y=4x^3+b{x}^2+3x-17$

where $b$ is a constant.

There is only one point on the curve where the gradient of the tangent at that point is zero.

Find the possible values of $b$.

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

$\mathrm{f}\left(x\right)=2x^3+p{x}^2+3x-16$

where $p$ is a constant.

Find all the possible values of $p$ for which the equation

$\mathrm{f}\text{'}\left(x\right)=0$

has at least one real solution.

A curve has the equation

$y=\frac{3}{8}{x}^{\frac{4}{3}}-12x^{\frac{1}{3}}$

Show that 

$\frac{dy}{dx}={ax}^{-\frac{2}{3}}\left(x+b\right)$

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 equation

$y={\left(\frac{1}{x}-\frac{1}{x\sqrt{x}}\right)}^{2}x>0$

Find $\frac{dy}{dx}$, writing your answer in the form

$\frac{\mathrm{d}y}{\mathrm{d}x}=a{x}^{k_1}+b{x}^{k_2}+c{x}^{k_3}$

where

• $a$, $b$ and $c$ are constants to be found

• $k_1$, $k_2$ and $k_3$ are terms of a descending arithmetic sequence to be found

A curve has equation 

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

Find $\frac{dy}{dx}$, writing the terms in your answer in ascending powers of $x$.

A curve has the equation

$y=a{x}^2+bx+c$

where $a$, $b$ and $c$ are non-zero constants.

• The curve passes through the point $\left(-1,4\right)$

• The gradient at the point $\left(2,7\right)$ is $7$

Find $a$, $b$ and $c$.