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Differentiation · Subtopic 8 of 10

Implicit Differentiation

Differentiating curves like $x^2+xy+y^2=12$ that can't easily be written as $y=\ldots$ : differentiate every term with respect to $x$ , remembering that $y$ terms pick up a factor $\dfrac{dy}{dx}$ .

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Section 1 of 14

Learning objectives

Differentiate terms in $y$ with respect to $x$ using the chain rule: $\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}$ .
Differentiate products such as $xy$ using the product rule.
Rearrange to find $\dfrac{dy}{dx}$ and evaluate it at a point.
Find tangents, normals and points with horizontal tangents.

Pause and think

The circle $x^2+y^2=25$ isn't a function of $x$ : most $x$ -values give two $y$ -values. Yet at $(3,4)$ it clearly has a gradient. How might you find it?

Section 2 of 14

Core idea

Start hereGradients without solving for y

The circle $x^2+y^2=25$ isn't a function $y=f(x)$ : each $x$ gives two $y$ 's. Solving for $y$ gives $\pm\sqrt{25-x^2}$ , messy. And some curves, like $x^3+xy+y^3=11$ , can't be solved for $y$ at all. Yet they have tangents. Implicit differentiation finds the gradient directly from the equation.

The big idea

Differentiate every term with respect to $x$ , treating $y$ as a function of $x$ . By the chain rule, $\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}$ ; products like $xy$ need the product rule. Then solve for $\frac{dy}{dx}$ .

1. The key step

\frac{d}{dx}(y^2)

$\frac{d}{dx}(y^2)=\frac{d}{dy}(y^2)\cdot\frac{dy}{dx}$
Chain rule: $y$ depends on $x$ .
$=2y\frac{dy}{dx}$
Differentiate $y^2$ with respect to $y$ .

Similarly

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

(product rule) and

\frac{d}{dx}e^y=e^y\frac{dy}{dx}

Worked exampleCircle $x^2+y^2=25$

Find $\frac{dy}{dx}$ and the gradient at $(3,4)$ .

Goal: Gradient.
Strategy: Differentiate term by term.

$2x+2y\frac{dy}{dx}=0$
$25$ is constant.
$\frac{dy}{dx}=-\frac xy$
Solve.
$(3,4):\ -\tfrac34$
Substitute.

Answer

-\frac xy

;

-\frac34

✓ CheckRadius gradient

\frac43

; product

-1

: tangent ⟂ radius ✓.

Gradient −x/y going round the circle

angle (°) = 10

On the right of the top half the tangent slopes down (

x>0

y>0

, so

-\tfrac xy<0

); on the left it slopes up. At the very top,

x=0

and it is flat.

2. Harder curves

Worked example $x^3+xy+y^3=11$ at $(1,2)$

Find the gradient.

Goal: A number.
Strategy: Differentiate; substitute before rearranging (easier).

$3x^2+\left(x\frac{dy}{dx}+y\right)+3y^2\frac{dy}{dx}=0$
Product rule on $xy$ .
$3+\frac{dy}{dx}+2+12\frac{dy}{dx}=0$
$x=1$ , $y=2$ .
$13\frac{dy}{dx}=-5\Rightarrow\frac{dy}{dx}=-\tfrac5{13}$
Solve.

Answer

-\frac5{13}

✓ Check

(1,2)

is on the curve:

1+2+8=11

✓.

Worked example $e^y+\sin x=yx^2$

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

Goal: General expression.
Strategy: Differentiate; collect $\frac{dy}{dx}$ terms; factorise.

$e^y\frac{dy}{dx}+\cos x=x^2\frac{dy}{dx}+2xy$
Product rule on $yx^2$ .
$\frac{dy}{dx}(e^y-x^2)=2xy-\cos x$
Collect.
$\frac{dy}{dx}=\frac{2xy-\cos x}{e^y-x^2}$
Divide.

Answer

\dfrac{2xy-\cos x}{e^y-x^2}

✓ CheckEach term was differentiated with respect to

x

✓.

Common misconceptionForgetting $\frac{dy}{dx}$

\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}

, not

3y^2

Key takeawayHorizontal and vertical tangents Horizontal: numerator of

\frac{dy}{dx}

0

. Vertical: denominator is

0

Pause and think

Find $\frac{dy}{dx}$ for $y^2=4x$ .

Summary

$y$ terms gain $\frac{dy}{dx}$ .
Product rule for mixed terms.
Collect and factorise.

Implicit: $y$ not written as $f(x)$ .

The tangent to x² + y² = 25 at (3, 4)

Differentiating both sides:

2x+2y\dfrac{dy}{dx}=0

, so

\dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac34

(3,4)

, without ever making

y

the subject.

HistoryDescartes' challenge. In 1638 René Descartes challenged Pierre de Fermat to find the tangent to the curve

x^3+y^3=3axy

, now called the folium of Descartes. Fermat solved it easily, to Descartes' annoyance. Implicit differentiation makes it a two-line exercise today.

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