1.10.5 Implicit and parametric differentiation (A-level only)

What you'll learn

How to differentiate a relation in which $y$ is not isolated.
Why differentiating a function of $y$ requires the chain rule.
How to find $dydx\frac{dy}{dx}$ when $x$ and $y$ are both defined using a parameter.
How to use derivatives to find gradients, stationary points and tangent equations.

Before You Start

You already know how to differentiate an explicit function, where $y$ is written directly in terms of $x$ , such as

y=x^3+2x.

You also need the chain rule. If $y$ depends on $x$ , then differentiating a function of $y$ with respect to $x$ introduces a factor of $dydx\frac{dy}{dx}$ . For example,

ddx(y3)=3y2dydx.\frac{d}{dx}\left(y^3\right)=3y^2\frac{dy}{dx}.

This is the key idea behind implicit differentiation.

Common Mistake

Forgetting the chain-rule factor

When differentiating a term involving $y$ with respect to $x$ , include $dydx\frac{dy}{dx}$ . For instance, $ddx(y2)=2ydydx\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ , not just $2 y$ .

Implicit Relations

An equation such as

x^2+y^2=25

describes a circle, but it does not initially give $y$ as a single function of $x$ . The variables are linked by a relation.

Definition

Implicit relation

An implicit relation is an equation connecting $x$ and $y$ without necessarily writing $y$ explicitly as a function of $x$ .

Although you could rearrange the circle equation to obtain $y=±25−x2y=\pm\sqrt{25-x^2}$ , this creates two branches. Implicit differentiation usually avoids that extra work.

Implicit Differentiation

To differentiate an implicit relation, differentiate every term with respect to $x$ . Treat $y$ as a function of $x$ , even though that function has not been written down.

Key Idea

Differentiate through y

Whenever differentiation acts on a function of $y$ , differentiate with respect to $y$ and then multiply by $dydx\frac{dy}{dx}$ .

Example

Finding the gradient of a circle

For the circle $x^2+y^2=25$ , find $dydx\frac{dy}{dx}$ and then find the gradient at $(3, 4)$ .

Differentiate both sides with respect to $x$ :
$2x+2ydydx=0.2x+2y\frac{dy}{dx}=0.$
Collect the term containing $dydx\frac{dy}{dx}$ and rearrange:
$2ydydx=−2x,dydx=−xy.\begin{aligned} 2y\frac{dy}{dx}&=-2x,\\ \frac{dy}{dx}&=-\frac{x}{y}. \end{aligned}$
Substitute $x = 3$ and $y = 4$ :
$dydx=−34.\frac{dy}{dx}=-\frac{3}{4}.$
Therefore, the tangent gradient at $(3, 4)$ is $−34-\frac34$ .

The radius to $(3, 4)$ has gradient $43\frac43$ , while the tangent has gradient $−34-\frac34$ . Their product is negative one, confirming that they are perpendicular.

Circle x squared plus y squared equals 25 with radius and tangent at P(3,4)

Products Containing x and y

An implicit relation may contain a product such as $x y$ . Because both factors vary with $x$ , use the product rule:

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

Here, the derivative of $x$ is one, while the derivative of $y$ is $dydx\frac{dy}{dx}$ .

Example

Differentiating a relation with a product

Find $dydx\frac{dy}{dx}$ if

x^2+xy+y^2=7.

Differentiate each term, using the product rule on $x y$ :
$2x+(xdydx+y)+2ydydx=0.2x+\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}=0.$
Group the terms containing $dydx\frac{dy}{dx}$ :
$(x+2y)dydx+2x+y=0.\left(x+2y\right)\frac{dy}{dx}+2x+y=0.$
Rearrange to make $dydx\frac{dy}{dx}$ the subject:
$dydx=−2x+yx+2y.\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$

Tip

Collect before dividing

After differentiating, move every term containing $dydx\frac{dy}{dx}$ to one side. Factor out $dydx\frac{dy}{dx}$ before dividing.

Functions of y

The chain rule applies to powers, exponentials and trigonometric functions of $y$ . For example,

ddx(sin⁡y)=cos⁡ydydx,ddx(ey)=eydydx,ddx(y)=12ydydx.\begin{aligned} \frac{d}{dx}(\sin y)&=\cos y\frac{dy}{dx},\\ \frac{d}{dx}(e^y)&=e^y\frac{dy}{dx},\\ \frac{d}{dx}\left(\sqrt{y}\right)&=\frac{1}{2\sqrt{y}}\frac{dy}{dx}. \end{aligned}

Example

Differentiating a trigonometric relation

Given

x^2+\sin y=1,

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

Differentiate both sides with respect to $x$ :
$2x+cos⁡ydydx=0.2x+\cos y\frac{dy}{dx}=0.$
Isolate the derivative term:
$cos⁡ydydx=−2x.\cos y\frac{dy}{dx}=-2x.$
Divide by $cos⁡y\cos y$ :
$dydx=−2xcos⁡y.\frac{dy}{dx}=-\frac{2x}{\cos y}.$

Tangents and Stationary Points

Once you have found $dydx\frac{dy}{dx}$ , you can use it exactly as you would for an explicit function.

A tangent is a straight line that has the same gradient as the curve at the point of contact. A stationary point is a point where the tangent is horizontal, so $dydx=0\frac{dy}{dx}=0$ .

Example

Finding a tangent to an implicit curve

For $x^2+xy+y^2=7$ , find the tangent at $(1, 2)$ .

Use the derivative already obtained:
$dydx=−2x+yx+2y.\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$
Substitute the coordinates:
$dydx∣(1,2)=−2(1)+21+2(2)=−45.\left.\frac{dy}{dx}\right|_{(1,2)} =-\frac{2(1)+2}{1+2(2)} =-\frac45.$
Use the point-gradient form $y-y_1=m(x-x_1)$ :
$y−2=−45(x−1).y-2=-\frac45(x-1).$

Common Mistake

When the derivative is undefined

If the denominator of your expression for $dydx\frac{dy}{dx}$ is zero while the numerator is non-zero, the curve may have a vertical tangent. Do not describe this as a stationary point.

Parametric Curves

Sometimes $x$ and $y$ are each given in terms of a third variable.

Definition

Parameter

A parameter is an extra variable, usually $t$ , that determines both coordinates of a point on a curve. Equations such as $x = f (t)$ and $y = g (t)$ are called parametric equations.

As $t$ changes, the point $(x, y)$ moves along the curve. To find its gradient, differentiate both coordinates with respect to $t$ .

Key Idea

Parametric derivative

Provided $dxdt≠0\frac{dx}{dt}\neq0$ ,

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

This follows from the chain rule:

dydt=dydxdxdt.\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}.

Example

Differentiating parametric equations

A curve is defined by

x=t2+1,y=t3−3t.x=t^2+1,\qquad y=t^3-3t.

Find $dydx\frac{dy}{dx}$ and its value when $t = 2$ .

Differentiate each coordinate with respect to $t$ :
$dxdt=2t,dydt=3t2−3.\frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=3t^2-3.$
Divide $dydt\frac{dy}{dt}$ by $dxdt\frac{dx}{dt}$ :
$dydx=3t2−32t=3(t2−1)2t.\frac{dy}{dx} =\frac{3t^2-3}{2t} =\frac{3(t^2-1)}{2t}.$
Substitute $t = 2$ :
$dydx=3(4−1)4=94.\frac{dy}{dx} =\frac{3(4-1)}{4} =\frac94.$

Common Mistake

Dividing in the wrong order

The correct order is $dydt÷dxdt\frac{dy}{dt}\div\frac{dx}{dt}$ . Writing the fractions in the opposite order gives $dxdy\frac{dx}{dy}$ .

Stationary Points on Parametric Curves

For a horizontal tangent, you normally need

dydt=0\frac{dy}{dt}=0

while also checking that $dxdt≠0\frac{dx}{dt}\neq0$ .

If $dxdt=0\frac{dx}{dt}=0$ , the formula for $dydx\frac{dy}{dx}$ involves division by zero and may indicate a vertical tangent or another special point requiring closer inspection.

Example

Finding a parametric stationary point

For

x=t2+1,y=t3−3t,x=t^2+1,\qquad y=t^3-3t,

find the points where the tangent is horizontal.

Set the numerator of $dydx\frac{dy}{dx}$ equal to zero:
$3t2−3=0⇒t2=1⇒t=±1.3t^2-3=0 \quad\Rightarrow\quad t^2=1 \quad\Rightarrow\quad t=\pm1.$
Check the denominator: $dxdt=2t\frac{dx}{dt}=2t$ , which is non-zero at both parameter values. Both give horizontal tangents.
Find the coordinates. When $t = 1$ , $(x, y) = (2, - 2)$ . When $t = - 1$ , $(x, y) = (2, 2)$ . Therefore, the stationary points are $(2, - 2)$ and $(2, 2)$ .

Exam technique

In the exam

For an implicit relation, differentiate every term with respect to $x$ and attach $dydx\frac{dy}{dx}$ whenever you differentiate a function of $y$ .
Use the product rule carefully for mixed terms such as $x y$ , then collect and factor all derivative terms.
For parametric equations, calculate $dydt\frac{dy}{dt}$ and $dxdt\frac{dx}{dt}$ separately before forming $dydx\frac{dy}{dx}$ , and check the denominator before identifying a stationary point.
Substitute both coordinates into an implicit derivative, but substitute the parameter value into a parametric derivative.

Self review

Check yourself

Can you find $dydx\frac{dy}{dx}$ for $x^3+y^3=6xy$ ?
For $x = 2 t + 1$ and $y=t^2-4t$ , when is the tangent horizontal?
What might it mean geometrically if $dxdt=0\frac{dx}{dt}=0$ but $dydt≠0\frac{dy}{dt}\neq0$ ?

What you'll learn

How to differentiate a relation in which $y$ is not isolated.
Why differentiating a function of $y$ requires the chain rule.
How to find $dydx\frac{dy}{dx}$ when $x$ and $y$ are both defined using a parameter.
How to use derivatives to find gradients, stationary points and tangent equations.

Before You Start

You already know how to differentiate an explicit function, where $y$ is written directly in terms of $x$ , such as

y=x^3+2x.

You also need the chain rule. If $y$ depends on $x$ , then differentiating a function of $y$ with respect to $x$ introduces a factor of $dydx\frac{dy}{dx}$ . For example,

ddx(y3)=3y2dydx.\frac{d}{dx}\left(y^3\right)=3y^2\frac{dy}{dx}.

This is the key idea behind implicit differentiation.

Common Mistake

Forgetting the chain-rule factor

When differentiating a term involving $y$ with respect to $x$ , include $dydx\frac{dy}{dx}$ . For instance, $ddx(y2)=2ydydx\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ , not just $2 y$ .

Implicit Relations

An equation such as

x^2+y^2=25

describes a circle, but it does not initially give $y$ as a single function of $x$ . The variables are linked by a relation.

Definition

Implicit relation

An implicit relation is an equation connecting $x$ and $y$ without necessarily writing $y$ explicitly as a function of $x$ .

Although you could rearrange the circle equation to obtain $y=±25−x2y=\pm\sqrt{25-x^2}$ , this creates two branches. Implicit differentiation usually avoids that extra work.

Implicit Differentiation

To differentiate an implicit relation, differentiate every term with respect to $x$ . Treat $y$ as a function of $x$ , even though that function has not been written down.

Key Idea

Differentiate through y

Whenever differentiation acts on a function of $y$ , differentiate with respect to $y$ and then multiply by $dydx\frac{dy}{dx}$ .

Example

Finding the gradient of a circle

For the circle $x^2+y^2=25$ , find $dydx\frac{dy}{dx}$ and then find the gradient at $(3, 4)$ .

Differentiate both sides with respect to $x$ :
$2x+2ydydx=0.2x+2y\frac{dy}{dx}=0.$
Collect the term containing $dydx\frac{dy}{dx}$ and rearrange:
$2ydydx=−2x,dydx=−xy.\begin{aligned} 2y\frac{dy}{dx}&=-2x,\\ \frac{dy}{dx}&=-\frac{x}{y}. \end{aligned}$
Substitute $x = 3$ and $y = 4$ :
$dydx=−34.\frac{dy}{dx}=-\frac{3}{4}.$
Therefore, the tangent gradient at $(3, 4)$ is $−34-\frac34$ .

The radius to $(3, 4)$ has gradient $43\frac43$ , while the tangent has gradient $−34-\frac34$ . Their product is negative one, confirming that they are perpendicular.

Circle x squared plus y squared equals 25 with radius and tangent at P(3,4)

Products Containing x and y

An implicit relation may contain a product such as $x y$ . Because both factors vary with $x$ , use the product rule:

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

Here, the derivative of $x$ is one, while the derivative of $y$ is $dydx\frac{dy}{dx}$ .

Example

Differentiating a relation with a product

Find $dydx\frac{dy}{dx}$ if

x^2+xy+y^2=7.

Differentiate each term, using the product rule on $x y$ :
$2x+(xdydx+y)+2ydydx=0.2x+\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}=0.$
Group the terms containing $dydx\frac{dy}{dx}$ :
$(x+2y)dydx+2x+y=0.\left(x+2y\right)\frac{dy}{dx}+2x+y=0.$
Rearrange to make $dydx\frac{dy}{dx}$ the subject:
$dydx=−2x+yx+2y.\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$

Tip

Collect before dividing

After differentiating, move every term containing $dydx\frac{dy}{dx}$ to one side. Factor out $dydx\frac{dy}{dx}$ before dividing.

Functions of y

The chain rule applies to powers, exponentials and trigonometric functions of $y$ . For example,

ddx(sin⁡y)=cos⁡ydydx,ddx(ey)=eydydx,ddx(y)=12ydydx.\begin{aligned} \frac{d}{dx}(\sin y)&=\cos y\frac{dy}{dx},\\ \frac{d}{dx}(e^y)&=e^y\frac{dy}{dx},\\ \frac{d}{dx}\left(\sqrt{y}\right)&=\frac{1}{2\sqrt{y}}\frac{dy}{dx}. \end{aligned}

Example

Differentiating a trigonometric relation

Given

x^2+\sin y=1,

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

Differentiate both sides with respect to $x$ :
$2x+cos⁡ydydx=0.2x+\cos y\frac{dy}{dx}=0.$
Isolate the derivative term:
$cos⁡ydydx=−2x.\cos y\frac{dy}{dx}=-2x.$
Divide by $cos⁡y\cos y$ :
$dydx=−2xcos⁡y.\frac{dy}{dx}=-\frac{2x}{\cos y}.$

Tangents and Stationary Points

Once you have found $dydx\frac{dy}{dx}$ , you can use it exactly as you would for an explicit function.

A tangent is a straight line that has the same gradient as the curve at the point of contact. A stationary point is a point where the tangent is horizontal, so $dydx=0\frac{dy}{dx}=0$ .

Example

Finding a tangent to an implicit curve

For $x^2+xy+y^2=7$ , find the tangent at $(1, 2)$ .

Use the derivative already obtained:
$dydx=−2x+yx+2y.\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$
Substitute the coordinates:
$dydx∣(1,2)=−2(1)+21+2(2)=−45.\left.\frac{dy}{dx}\right|_{(1,2)} =-\frac{2(1)+2}{1+2(2)} =-\frac45.$
Use the point-gradient form $y-y_1=m(x-x_1)$ :
$y−2=−45(x−1).y-2=-\frac45(x-1).$

Common Mistake

When the derivative is undefined

If the denominator of your expression for $dydx\frac{dy}{dx}$ is zero while the numerator is non-zero, the curve may have a vertical tangent. Do not describe this as a stationary point.

Parametric Curves

Sometimes $x$ and $y$ are each given in terms of a third variable.

Definition

Parameter

A parameter is an extra variable, usually $t$ , that determines both coordinates of a point on a curve. Equations such as $x = f (t)$ and $y = g (t)$ are called parametric equations.

As $t$ changes, the point $(x, y)$ moves along the curve. To find its gradient, differentiate both coordinates with respect to $t$ .

Key Idea

Parametric derivative

Provided $dxdt≠0\frac{dx}{dt}\neq0$ ,

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

This follows from the chain rule:

dydt=dydxdxdt.\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}.

Example

Differentiating parametric equations

A curve is defined by

x=t2+1,y=t3−3t.x=t^2+1,\qquad y=t^3-3t.

Find $dydx\frac{dy}{dx}$ and its value when $t = 2$ .

Differentiate each coordinate with respect to $t$ :
$dxdt=2t,dydt=3t2−3.\frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=3t^2-3.$
Divide $dydt\frac{dy}{dt}$ by $dxdt\frac{dx}{dt}$ :
$dydx=3t2−32t=3(t2−1)2t.\frac{dy}{dx} =\frac{3t^2-3}{2t} =\frac{3(t^2-1)}{2t}.$
Substitute $t = 2$ :
$dydx=3(4−1)4=94.\frac{dy}{dx} =\frac{3(4-1)}{4} =\frac94.$

Common Mistake

Dividing in the wrong order

The correct order is $dydt÷dxdt\frac{dy}{dt}\div\frac{dx}{dt}$ . Writing the fractions in the opposite order gives $dxdy\frac{dx}{dy}$ .

Stationary Points on Parametric Curves

For a horizontal tangent, you normally need

dydt=0\frac{dy}{dt}=0

while also checking that $dxdt≠0\frac{dx}{dt}\neq0$ .

If $dxdt=0\frac{dx}{dt}=0$ , the formula for $dydx\frac{dy}{dx}$ involves division by zero and may indicate a vertical tangent or another special point requiring closer inspection.

Example

Finding a parametric stationary point

For

x=t2+1,y=t3−3t,x=t^2+1,\qquad y=t^3-3t,

find the points where the tangent is horizontal.

Set the numerator of $dydx\frac{dy}{dx}$ equal to zero:
$3t2−3=0⇒t2=1⇒t=±1.3t^2-3=0 \quad\Rightarrow\quad t^2=1 \quad\Rightarrow\quad t=\pm1.$
Check the denominator: $dxdt=2t\frac{dx}{dt}=2t$ , which is non-zero at both parameter values. Both give horizontal tangents.
Find the coordinates. When $t = 1$ , $(x, y) = (2, - 2)$ . When $t = - 1$ , $(x, y) = (2, 2)$ . Therefore, the stationary points are $(2, - 2)$ and $(2, 2)$ .

Exam technique

In the exam

For an implicit relation, differentiate every term with respect to $x$ and attach $dydx\frac{dy}{dx}$ whenever you differentiate a function of $y$ .
Use the product rule carefully for mixed terms such as $x y$ , then collect and factor all derivative terms.
For parametric equations, calculate $dydt\frac{dy}{dt}$ and $dxdt\frac{dx}{dt}$ separately before forming $dydx\frac{dy}{dx}$ , and check the denominator before identifying a stationary point.
Substitute both coordinates into an implicit derivative, but substitute the parameter value into a parametric derivative.

Self review

Check yourself

Can you find $dydx\frac{dy}{dx}$ for $x^3+y^3=6xy$ ?
For $x = 2 t + 1$ and $y=t^2-4t$ , when is the tangent horizontal?
What might it mean geometrically if $dxdt=0\frac{dx}{dt}=0$ but $dydt≠0\frac{dy}{dt}\neq0$ ?

What you'll learn

Before You Start

Implicit Relations

Implicit Differentiation

Products Containing x and y

Functions of y

Tangents and Stationary Points

Parametric Curves

Stationary Points on Parametric Curves

Review 1.10.5 Implicit and parametric differentiation (A-level only) by teaching Genie

1.10.5 Implicit and parametric differentiation (A-level only)

What you'll learn

Before You Start

Implicit Relations

Implicit Differentiation

Products Containing x and y

Functions of y

Tangents and Stationary Points

Parametric Curves

Stationary Points on Parametric Curves

Review 1.10.5 Implicit and parametric differentiation (A-level only) by teaching Genie

1.10.5 Implicit and parametric differentiation (A-level only)

What you'll learn

Before You Start

Implicit Relations

Implicit Differentiation

Products Containing x and y

Functions of y

Tangents and Stationary Points

Parametric Curves

Stationary Points on Parametric Curves

Review 1.10.5 Implicit and parametric differentiation (A-level only) by teaching Genie

1.10.5 Implicit and parametric differentiation (A-level only) Revision Guide

1.10.5 Implicit and parametric differentiation (A-level only)

What you'll learn

Before You Start

Implicit Relations

Implicit Differentiation

Products Containing x and y

Functions of y

Tangents and Stationary Points

Parametric Curves

Stationary Points on Parametric Curves

Review 1.10.5 Implicit and parametric differentiation (A-level only) by teaching Genie

1.10.5 Implicit and parametric differentiation (A-level only) Revision Guide