Differentiation

What you'll learn

What $dydx\frac{dy}{dx}$ means: the gradient of a curve at a point.
How to use the power, chain, product and quotient rules.
How to find tangents, normals and stationary points.
How to deal with implicit and parametric curves.

The derivative: gradient at a point

A straight line has one constant gradient. A curve has a changing gradient, so we need a way to find the gradient at one particular point.

Definition

Derivative

The derivative of $y$ with respect to $x$ , written $dydx\frac{dy}{dx}$ , is the instantaneous rate of change of $y$ as $x$ changes. If $y = f (x)$ , the derivative can also be written as $f^{'} (x)$ .

From first principles, the derivative is defined by

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

This means: find the gradient of a tiny chord, then let the chord shrink until it becomes a tangent.

A secant chord between two nearby points on a curve approaches the tangent as the horizontal gap h tends to zero.

Example

Differentiating from first principles

Let $f(x)=x^2+5x$ . Show that $f^{'} (x) = 2 x + 5$ .

Start with $f (x + h)$ :
$f(x+h)=(x+h)^2+5(x+h)$
Form the difference quotient and simplify:
$f(x+h)−f(x)h=x2+2xh+h2+5x+5h−(x2+5x)h=2x+h+5\frac{f(x+h)-f(x)}{h} = \frac{x^2+2xh+h^2+5x+5h-(x^2+5x)}{h} = 2x+h+5$
Let $h→0h\to 0$ :
$f^{'} (x) = 2 x + 5$

Standard derivatives and rewriting first

Before differentiating, it often helps to rewrite roots and fractions using indices:

$x=x12\sqrt{x}=x^{\frac12}$
$1x2=x−2\frac{1}{x^2}=x^{-2}$
$22x+1=2(2x+1)−12\frac{2}{\sqrt{2x+1}}=2(2x+1)^{-\frac12}$

The main standard results are:

Power rule: if $y=x^n$ , then $dydx=nxn−1\frac{dy}{dx}=nx^{n-1}$ .
$ddx(ex)=ex\frac{d}{dx}(e^x)=e^x$
$ddx(ln⁡x)=1x\frac{d}{dx}(\ln x)=\frac1x$ , for $x > 0$
$ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x)=\cos x$
$ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x)=-\sin x$
$ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x)=\sec^2 x$
$ddx(ax)=axln⁡a\frac{d}{dx}(a^x)=a^x\ln a$ , for $a > 0$

Common Mistake

Radians only

The standard trigonometric derivatives assume angles are measured in radians. This is why A-Level differentiation questions involving trig usually work in radians.

Example

Finding first and second derivatives

Let $f(x)=\ln x+e^x+4x^3$ . Find $f^{'} (x)$ and $f^{''} (x)$ .

Differentiate each term separately:
$f′(x)=1x+ex+12x2f'(x)=\frac1x+e^x+12x^2$
Differentiate again to get the second derivative:
$f′′(x)=−1x2+ex+24xf''(x)=-\frac{1}{x^2}+e^x+24x$

Definition

Second derivative

The second derivative, written $d2ydx2\frac{d^2y}{dx^2}$ or $f^{''} (x)$ , is the derivative of the derivative. It tells you how the gradient is changing.

The chain rule

A composite function is a function inside another function, such as $x^2+4)^3$ or $e^{x^2}$ .

The chain rule treats a composite function as an inside function passed into an outside function.

Key Idea

Chain rule

Differentiate the outside function, keep the inside unchanged, then multiply by the derivative of the inside.

If $y = f (g (x))$ , then

dydx=f′(g(x))g′(x)\frac{dy}{dx}=f'(g(x))g'(x)

Example

Using the chain rule

Differentiate $y=(x^2+4)^3$ and $z=6+e^{x^2}$ .

For $y=(x^2+4)^3$ , the outside is “cube” and the inside is $x^2+4$ :
$dydx=3(x2+4)2⋅2x=6x(x2+4)2\frac{dy}{dx}=3(x^2+4)^2\cdot 2x=6x(x^2+4)^2$
For $z=6+e^{x^2}$ , the constant differentiates to zero and the inside of the exponential is $x^2$ :
$dzdx=ex2⋅2x=2xex2\frac{dz}{dx}=e^{x^2}\cdot 2x=2xe^{x^2}$

Common Mistake

Forgetting the inside derivative

For $e^{x^2}$ , the derivative is $2xe^{x^2}$ , not just $e^{x^2}$ . The extra $2 x$ comes from differentiating the inside.

Product and quotient rules

Use the product rule when two expressions involving $x$ are multiplied.

ddx(uv)=u′v+uv′\frac{d}{dx}(uv)=u'v+uv'

Use the quotient rule when one expression involving $x$ is divided by another.

ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}

Here, $u$ and $v$ are functions of $x$ , and $v≠0v\neq 0$ .

Example

Product and quotient rule practice

Differentiate $y=2xe^x$ and $z=ln⁡(x2+1)2x+3z=\frac{\ln(x^2+1)}{2x+3}$ .

For $y=2xe^x$ , take $u = 2 x$ and $v=e^x$ :
$dydx=2ex+2xex=2ex(x+1)\frac{dy}{dx}=2e^x+2xe^x=2e^x(x+1)$
For $z=ln⁡(x2+1)2x+3z=\frac{\ln(x^2+1)}{2x+3}$ , take $u=\ln(x^2+1)$ and $v = 2 x + 3$ :
$u′=2xx2+1,v′=2u'=\frac{2x}{x^2+1}, \qquad v'=2$
Substitute into the quotient rule:
$dzdx=2xx2+1(2x+3)−2ln⁡(x2+1)(2x+3)2\frac{dz}{dx} = \frac{\frac{2x}{x^2+1}(2x+3)-2\ln(x^2+1)}{(2x+3)^2}$

Tip

Simplify first if you can

If a quotient can be simplified algebraically before differentiating, do that first. It often saves time and reduces mistakes.

Tangents and normals

A tangent is a straight line that has the same gradient as the curve at a point.

A normal is a straight line perpendicular to the tangent. If the tangent gradient is $m$ , then the normal gradient is $−1m-\frac1m$ , as long as $m≠0m\neq 0$ .

At a point on a curve, the normal is perpendicular to the tangent and has the negative reciprocal gradient.

The equation of a straight line through $x_1,y_1)$ with gradient $m$ is

y-y_1=m(x-x_1)

Example

Finding an equation of a normal

The point $P$ lies on the curve $y=42x+1y=\frac{4}{2x+1}$ with $x$ -coordinate 1. Find an equation of the normal at $P$ .

The curve y = 4/(2x+1) with the point P at x = 1 and the normal line through P.

Find the coordinates of $P$ :
$y=42(1)+1=43y=\frac{4}{2(1)+1}=\frac43$
Differentiate by rewriting the curve as $y=4(2x+1)^{-1}$ :
$dydx=4(−1)(2x+1)−2⋅2=−8(2x+1)2\frac{dy}{dx}=4(-1)(2x+1)^{-2}\cdot 2=-\frac{8}{(2x+1)^2}$
Find the tangent gradient at $x = 1$ :
$mtangent=−89m_{\text{tangent}}=-\frac{8}{9}$
Use the negative reciprocal for the normal gradient:
$mnormal=98m_{\text{normal}}=\frac98$
Use $y-y_1=m(x-x_1)$ :
$y−43=98(x−1)y-\frac43=\frac98(x-1)$

Stationary points and nature

Definition

Stationary point

A stationary point is a point on a curve where $dydx=0\frac{dy}{dx}=0$ . It could be a local maximum, a local minimum, or a stationary point of inflection.

The three main types of stationary point all have a horizontal tangent but different local behaviour.

To classify stationary points:

If $d2ydx2>0\frac{d^2y}{dx^2}>0$ , the point is a local minimum.
If $d2ydx2<0\frac{d^2y}{dx^2}<0$ , the point is a local maximum.
If $d2ydx2=0\frac{d^2y}{dx^2}=0$ , the test is inconclusive, so check the sign of $dydx\frac{dy}{dx}$ around the point.

Example

Finding stationary points and their nature

Find the stationary points of $y=x(x-2)^3$ and determine their nature.

The curve y = x(x − 2)^3 has a local minimum at x = 1/2 and a stationary point of inflection at x = 2.

Differentiate using the product rule:
$dydx=(x−2)3+3x(x−2)2=(x−2)2(4x−2)\frac{dy}{dx}=(x-2)^3+3x(x-2)^2=(x-2)^2(4x-2)$
Set the derivative equal to zero:
$x-2)^2(4x-2)=0$
Solve for $x$ :
$\quad \text{or} \quad x=\frac12$
Find the corresponding $y$ -coordinates:
$\qquad y\left(\frac12\right)=\frac12\left(-\frac32\right)^3=-\frac{27}{16}$
Differentiate again:
$d2ydx2=12(x−2)(x−1)\frac{d^2y}{dx^2}=12(x-2)(x-1)$
At $x=12x=\frac12$ , $d2ydx2>0\frac{d^2y}{dx^2}>0$ , so $(12,−2716)\left(\frac12,-\frac{27}{16}\right)$ is a local minimum.
At $x = 2$ , the second derivative test is inconclusive. Since $dydx=(x−2)2(4x−2)\frac{dy}{dx}=(x-2)^2(4x-2)$ is positive on both sides of $x = 2$ , $(2, 0)$ is a stationary point of inflection.

Implicit differentiation

An implicit equation links $x$ and $y$ without necessarily making $y$ the subject, such as $x^2+xy+2y^2=23$ .

When differentiating terms involving $y$ , remember that $y$ is a function of $x$ .

For example:

ddx(y2)=2ydydx\frac{d}{dx}(y^2)=2y\frac{dy}{dx}

and, using the product rule,

ddx(xy)=xdydx+y\frac{d}{dx}(xy)=x\frac{dy}{dx}+y

Example

Implicit differentiation and a normal

A curve has equation $x^2+xy+2y^2=23$ . Find an equation of the normal at $(3, 2)$ .

The implicit curve with its tangent and normal at the point (3, 2).

Differentiate each term with respect to $x$ :
$2x+(xdydx+y)+4ydydx=02x+\left(x\frac{dy}{dx}+y\right)+4y\frac{dy}{dx}=0$
Collect the $dydx\frac{dy}{dx}$ terms:
$(x+4y)dydx=−(2x+y)(x+4y)\frac{dy}{dx}=-(2x+y)$
Substitute $(3, 2)$ to find the tangent gradient:
$mtangent=−2(3)+23+4(2)=−811m_{\text{tangent}}=-\frac{2(3)+2}{3+4(2)}=-\frac{8}{11}$
Find the normal gradient:
$mnormal=118m_{\text{normal}}=\frac{11}{8}$
Use the line equation:
$y−2=118(x−3)y-2=\frac{11}{8}(x-3)$

Parametric differentiation

A curve is given parametrically when both $x$ and $y$ are written in terms of another variable, usually $t$ . This variable is called a parameter.

If $x = x (t)$ and $y = y (t)$ , then

dydx=dydtdxdt\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

provided $dxdt≠0\frac{dx}{dt}\neq 0$ .

Example

A parametric tangent

A curve has parametric equations $x=3cos⁡tx=3\cos t$ and $y=2sin⁡ty=2\sin t$ . Find where the tangent at $t=π3t=\frac{\pi}{3}$ cuts the $y$ -axis.

The parametric curve x = 3 cos t, y = 2 sin t is an ellipse, with a tangent at t = π/3 meeting the y-axis.

Differentiate both parametric equations:
$dxdt=−3sin⁡t,dydt=2cos⁡t\frac{dx}{dt}=-3\sin t, \qquad \frac{dy}{dt}=2\cos t$
Divide to find $dydx\frac{dy}{dx}$ :
$dydx=2cos⁡t−3sin⁡t=−23cot⁡t\frac{dy}{dx}=\frac{2\cos t}{-3\sin t}=-\frac23\cot t$
At $t=π3t=\frac{\pi}{3}$ , find the point and gradient:
$(x,y)=(32,3),m=−233(x,y)=\left(\frac32,\sqrt3\right), \qquad m=-\frac{2}{3\sqrt3}$
Write the tangent equation:
$y−3=−233(x−32)y-\sqrt3=-\frac{2}{3\sqrt3}\left(x-\frac32\right)$
Set $x = 0$ to find the intercept on the $y$ -axis:
$y=3+13=433y=\sqrt3+\frac{1}{\sqrt3}=\frac{4\sqrt3}{3}$

Exam technique

In the exam

Rewrite roots and fractions as powers before choosing a rule.
For tangents and normals, always find both the point and the gradient.
For stationary points, solve $dydx=0\frac{dy}{dx}=0$ and then justify the nature using $d2ydx2\frac{d^2y}{dx^2}$ or a sign test.
For implicit differentiation, collect all $dydx\frac{dy}{dx}$ terms on one side before factorising.
For parametric curves, find $dydx\frac{dy}{dx}$ by dividing $dydt\frac{dy}{dt}$ by $dxdt\frac{dx}{dt}$ , not the other way round.

Self review

Check yourself

Can you explain why differentiating $e^{x^2}$ gives an extra factor of $2 x$ ?
When finding a normal, why do you need the negative reciprocal of the tangent gradient?
How would you prove that a function is decreasing over its whole domain?

What you'll learn

What $dydx\frac{dy}{dx}$ means: the gradient of a curve at a point.
How to use the power, chain, product and quotient rules.
How to find tangents, normals and stationary points.
How to deal with implicit and parametric curves.

The derivative: gradient at a point

A straight line has one constant gradient. A curve has a changing gradient, so we need a way to find the gradient at one particular point.

Definition

Derivative

From first principles, the derivative is defined by

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

This means: find the gradient of a tiny chord, then let the chord shrink until it becomes a tangent.

A secant chord between two nearby points on a curve approaches the tangent as the horizontal gap h tends to zero.

Example

Differentiating from first principles

Let $f(x)=x^2+5x$ . Show that $f^{'} (x) = 2 x + 5$ .

Start with $f (x + h)$ :
$f(x+h)=(x+h)^2+5(x+h)$
Form the difference quotient and simplify:
$f(x+h)−f(x)h=x2+2xh+h2+5x+5h−(x2+5x)h=2x+h+5\frac{f(x+h)-f(x)}{h} = \frac{x^2+2xh+h^2+5x+5h-(x^2+5x)}{h} = 2x+h+5$
Let $h→0h\to 0$ :
$f^{'} (x) = 2 x + 5$

Standard derivatives and rewriting first

Before differentiating, it often helps to rewrite roots and fractions using indices:

$x=x12\sqrt{x}=x^{\frac12}$
$1x2=x−2\frac{1}{x^2}=x^{-2}$
$22x+1=2(2x+1)−12\frac{2}{\sqrt{2x+1}}=2(2x+1)^{-\frac12}$

The main standard results are:

Power rule: if $y=x^n$ , then $dydx=nxn−1\frac{dy}{dx}=nx^{n-1}$ .
$ddx(ex)=ex\frac{d}{dx}(e^x)=e^x$
$ddx(ln⁡x)=1x\frac{d}{dx}(\ln x)=\frac1x$ , for $x > 0$
$ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x)=\cos x$
$ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x)=-\sin x$
$ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x)=\sec^2 x$
$ddx(ax)=axln⁡a\frac{d}{dx}(a^x)=a^x\ln a$ , for $a > 0$

Common Mistake

Radians only

The standard trigonometric derivatives assume angles are measured in radians. This is why A-Level differentiation questions involving trig usually work in radians.

Example

Finding first and second derivatives

Let $f(x)=\ln x+e^x+4x^3$ . Find $f^{'} (x)$ and $f^{''} (x)$ .

Differentiate each term separately:
$f′(x)=1x+ex+12x2f'(x)=\frac1x+e^x+12x^2$
Differentiate again to get the second derivative:
$f′′(x)=−1x2+ex+24xf''(x)=-\frac{1}{x^2}+e^x+24x$

Definition

Second derivative

The second derivative, written $d2ydx2\frac{d^2y}{dx^2}$ or $f^{''} (x)$ , is the derivative of the derivative. It tells you how the gradient is changing.

The chain rule

A composite function is a function inside another function, such as $x^2+4)^3$ or $e^{x^2}$ .

The chain rule treats a composite function as an inside function passed into an outside function.

Key Idea

Chain rule

Differentiate the outside function, keep the inside unchanged, then multiply by the derivative of the inside.

If $y = f (g (x))$ , then

dydx=f′(g(x))g′(x)\frac{dy}{dx}=f'(g(x))g'(x)

Example

Using the chain rule

Differentiate $y=(x^2+4)^3$ and $z=6+e^{x^2}$ .

For $y=(x^2+4)^3$ , the outside is “cube” and the inside is $x^2+4$ :
$dydx=3(x2+4)2⋅2x=6x(x2+4)2\frac{dy}{dx}=3(x^2+4)^2\cdot 2x=6x(x^2+4)^2$
For $z=6+e^{x^2}$ , the constant differentiates to zero and the inside of the exponential is $x^2$ :
$dzdx=ex2⋅2x=2xex2\frac{dz}{dx}=e^{x^2}\cdot 2x=2xe^{x^2}$

Common Mistake

Forgetting the inside derivative

For $e^{x^2}$ , the derivative is $2xe^{x^2}$ , not just $e^{x^2}$ . The extra $2 x$ comes from differentiating the inside.

Product and quotient rules

Use the product rule when two expressions involving $x$ are multiplied.

ddx(uv)=u′v+uv′\frac{d}{dx}(uv)=u'v+uv'

Use the quotient rule when one expression involving $x$ is divided by another.

ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}

Here, $u$ and $v$ are functions of $x$ , and $v≠0v\neq 0$ .

Example

Product and quotient rule practice

Differentiate $y=2xe^x$ and $z=ln⁡(x2+1)2x+3z=\frac{\ln(x^2+1)}{2x+3}$ .

For $y=2xe^x$ , take $u = 2 x$ and $v=e^x$ :
$dydx=2ex+2xex=2ex(x+1)\frac{dy}{dx}=2e^x+2xe^x=2e^x(x+1)$
For $z=ln⁡(x2+1)2x+3z=\frac{\ln(x^2+1)}{2x+3}$ , take $u=\ln(x^2+1)$ and $v = 2 x + 3$ :
$u′=2xx2+1,v′=2u'=\frac{2x}{x^2+1}, \qquad v'=2$
Substitute into the quotient rule:
$dzdx=2xx2+1(2x+3)−2ln⁡(x2+1)(2x+3)2\frac{dz}{dx} = \frac{\frac{2x}{x^2+1}(2x+3)-2\ln(x^2+1)}{(2x+3)^2}$

Tip

Simplify first if you can

If a quotient can be simplified algebraically before differentiating, do that first. It often saves time and reduces mistakes.

Tangents and normals

A tangent is a straight line that has the same gradient as the curve at a point.

A normal is a straight line perpendicular to the tangent. If the tangent gradient is $m$ , then the normal gradient is $−1m-\frac1m$ , as long as $m≠0m\neq 0$ .

At a point on a curve, the normal is perpendicular to the tangent and has the negative reciprocal gradient.

The equation of a straight line through $x_1,y_1)$ with gradient $m$ is

y-y_1=m(x-x_1)

Example

Finding an equation of a normal

The point $P$ lies on the curve $y=42x+1y=\frac{4}{2x+1}$ with $x$ -coordinate 1. Find an equation of the normal at $P$ .

The curve y = 4/(2x+1) with the point P at x = 1 and the normal line through P.

Find the coordinates of $P$ :
$y=42(1)+1=43y=\frac{4}{2(1)+1}=\frac43$
Differentiate by rewriting the curve as $y=4(2x+1)^{-1}$ :
$dydx=4(−1)(2x+1)−2⋅2=−8(2x+1)2\frac{dy}{dx}=4(-1)(2x+1)^{-2}\cdot 2=-\frac{8}{(2x+1)^2}$
Find the tangent gradient at $x = 1$ :
$mtangent=−89m_{\text{tangent}}=-\frac{8}{9}$
Use the negative reciprocal for the normal gradient:
$mnormal=98m_{\text{normal}}=\frac98$
Use $y-y_1=m(x-x_1)$ :
$y−43=98(x−1)y-\frac43=\frac98(x-1)$

Stationary points and nature

Definition

Stationary point

A stationary point is a point on a curve where $dydx=0\frac{dy}{dx}=0$ . It could be a local maximum, a local minimum, or a stationary point of inflection.

The three main types of stationary point all have a horizontal tangent but different local behaviour.

To classify stationary points:

If $d2ydx2>0\frac{d^2y}{dx^2}>0$ , the point is a local minimum.
If $d2ydx2<0\frac{d^2y}{dx^2}<0$ , the point is a local maximum.
If $d2ydx2=0\frac{d^2y}{dx^2}=0$ , the test is inconclusive, so check the sign of $dydx\frac{dy}{dx}$ around the point.

Example

Finding stationary points and their nature

Find the stationary points of $y=x(x-2)^3$ and determine their nature.

The curve y = x(x − 2)^3 has a local minimum at x = 1/2 and a stationary point of inflection at x = 2.

Differentiate using the product rule:
$dydx=(x−2)3+3x(x−2)2=(x−2)2(4x−2)\frac{dy}{dx}=(x-2)^3+3x(x-2)^2=(x-2)^2(4x-2)$
Set the derivative equal to zero:
$x-2)^2(4x-2)=0$
Solve for $x$ :
$\quad \text{or} \quad x=\frac12$
Find the corresponding $y$ -coordinates:
$\qquad y\left(\frac12\right)=\frac12\left(-\frac32\right)^3=-\frac{27}{16}$
Differentiate again:
$d2ydx2=12(x−2)(x−1)\frac{d^2y}{dx^2}=12(x-2)(x-1)$
At $x=12x=\frac12$ , $d2ydx2>0\frac{d^2y}{dx^2}>0$ , so $(12,−2716)\left(\frac12,-\frac{27}{16}\right)$ is a local minimum.
At $x = 2$ , the second derivative test is inconclusive. Since $dydx=(x−2)2(4x−2)\frac{dy}{dx}=(x-2)^2(4x-2)$ is positive on both sides of $x = 2$ , $(2, 0)$ is a stationary point of inflection.

Implicit differentiation

An implicit equation links $x$ and $y$ without necessarily making $y$ the subject, such as $x^2+xy+2y^2=23$ .

When differentiating terms involving $y$ , remember that $y$ is a function of $x$ .

For example:

ddx(y2)=2ydydx\frac{d}{dx}(y^2)=2y\frac{dy}{dx}

and, using the product rule,

ddx(xy)=xdydx+y\frac{d}{dx}(xy)=x\frac{dy}{dx}+y

Example

Implicit differentiation and a normal

A curve has equation $x^2+xy+2y^2=23$ . Find an equation of the normal at $(3, 2)$ .

The implicit curve with its tangent and normal at the point (3, 2).

Differentiate each term with respect to $x$ :
$2x+(xdydx+y)+4ydydx=02x+\left(x\frac{dy}{dx}+y\right)+4y\frac{dy}{dx}=0$
Collect the $dydx\frac{dy}{dx}$ terms:
$(x+4y)dydx=−(2x+y)(x+4y)\frac{dy}{dx}=-(2x+y)$
Substitute $(3, 2)$ to find the tangent gradient:
$mtangent=−2(3)+23+4(2)=−811m_{\text{tangent}}=-\frac{2(3)+2}{3+4(2)}=-\frac{8}{11}$
Find the normal gradient:
$mnormal=118m_{\text{normal}}=\frac{11}{8}$
Use the line equation:
$y−2=118(x−3)y-2=\frac{11}{8}(x-3)$

Parametric differentiation

A curve is given parametrically when both $x$ and $y$ are written in terms of another variable, usually $t$ . This variable is called a parameter.

If $x = x (t)$ and $y = y (t)$ , then

dydx=dydtdxdt\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

provided $dxdt≠0\frac{dx}{dt}\neq 0$ .

Example

A parametric tangent

A curve has parametric equations $x=3cos⁡tx=3\cos t$ and $y=2sin⁡ty=2\sin t$ . Find where the tangent at $t=π3t=\frac{\pi}{3}$ cuts the $y$ -axis.

The parametric curve x = 3 cos t, y = 2 sin t is an ellipse, with a tangent at t = π/3 meeting the y-axis.

Differentiate both parametric equations:
$dxdt=−3sin⁡t,dydt=2cos⁡t\frac{dx}{dt}=-3\sin t, \qquad \frac{dy}{dt}=2\cos t$
Divide to find $dydx\frac{dy}{dx}$ :
$dydx=2cos⁡t−3sin⁡t=−23cot⁡t\frac{dy}{dx}=\frac{2\cos t}{-3\sin t}=-\frac23\cot t$
At $t=π3t=\frac{\pi}{3}$ , find the point and gradient:
$(x,y)=(32,3),m=−233(x,y)=\left(\frac32,\sqrt3\right), \qquad m=-\frac{2}{3\sqrt3}$
Write the tangent equation:
$y−3=−233(x−32)y-\sqrt3=-\frac{2}{3\sqrt3}\left(x-\frac32\right)$
Set $x = 0$ to find the intercept on the $y$ -axis:
$y=3+13=433y=\sqrt3+\frac{1}{\sqrt3}=\frac{4\sqrt3}{3}$

Exam technique

In the exam

Rewrite roots and fractions as powers before choosing a rule.
For tangents and normals, always find both the point and the gradient.
For stationary points, solve $dydx=0\frac{dy}{dx}=0$ and then justify the nature using $d2ydx2\frac{d^2y}{dx^2}$ or a sign test.
For implicit differentiation, collect all $dydx\frac{dy}{dx}$ terms on one side before factorising.
For parametric curves, find $dydx\frac{dy}{dx}$ by dividing $dydt\frac{dy}{dt}$ by $dxdt\frac{dx}{dt}$ , not the other way round.

Self review

Check yourself

Can you explain why differentiating $e^{x^2}$ gives an extra factor of $2 x$ ?
When finding a normal, why do you need the negative reciprocal of the tangent gradient?
How would you prove that a function is decreasing over its whole domain?

What you'll learn

The derivative: gradient at a point

Standard derivatives and rewriting first

The chain rule

Product and quotient rules

Tangents and normals

Stationary points and nature

Implicit differentiation

Parametric differentiation

Recap questions

Review Differentiation by teaching Genie

Remember key concepts with flashcards

Differentiation

What you'll learn

The derivative: gradient at a point

Standard derivatives and rewriting first

The chain rule

Product and quotient rules

Tangents and normals

Stationary points and nature

Implicit differentiation

Parametric differentiation

Recap questions

Review Differentiation by teaching Genie

Remember key concepts with flashcards

Differentiation

What you'll learn

The derivative: gradient at a point

Standard derivatives and rewriting first

The chain rule

Product and quotient rules

Tangents and normals

Stationary points and nature

Implicit differentiation

Parametric differentiation

Recap questions

Review Differentiation by teaching Genie

Remember key concepts with flashcards

Differentiation Revision Guide

Differentiation

What you'll learn

The derivative: gradient at a point

Standard derivatives and rewriting first

The chain rule

Product and quotient rules

Tangents and normals

Stationary points and nature

Implicit differentiation

Parametric differentiation

Recap questions

Review Differentiation by teaching Genie

Remember key concepts with flashcards

Differentiation Revision Guide