Differentiation (Edexcel International AS Maths: Pure 1): Exam Questions

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:

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

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

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

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

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

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

Given that the equation ${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=0$ 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 $\left(-1,13\right),$ and a gradient of $-3$ at the point $\left(1,3\right)$. 

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 $P\left(2,9\right)$ 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 $P\left(1,1\right)$  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

 $\frac{dy}{dx}$

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

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

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

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

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

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

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

Show that $\frac{dy}{dx}=a{x}^{-\frac{2}{3}}\left(x+b\right),$ 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 the equation $y=4x^3+b{x}^2+3x-17,$ 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 $y=a{x}^2+bx+c$ has a gradient of 8 at the point $\left(-2,0\right)$, and a gradient of $-10$ at the point $\left(1,-3\right)$.  Find the values of $a,b$ and $c$.

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

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

Find the coordinates of $Q$.

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

$\frac{dy}{dx}$

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

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

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

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

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

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

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

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

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

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 $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{n}}-\mathrm{x},$ $n\in \mathrm{\mathbb{N}},n\ge 2.$ Determine the relationship between the value of $n$ and the number of real solutions to the equation ${\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=0$.

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  $y=a{x}^2+bx+c$ passes through the point $\left(-1,4\right)$.  At the point $\left(2,7\right)$ 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 $y=\mathrm{f}\left(\mathrm{x}\right)$, where

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

$P$ 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 $\left(2^a,2^b\right)$, 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

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