Implicit differentiation | Year 12

01 / LEARN

The idea, step by step

Treat y as a function of x

An implicit equation describes a relation without isolating one variable. In $x^{2} + y^{2} = 25$ , a vertical line usually meets two points, so solving for $y$ introduces separate branches. Implicit differentiation finds a local derivative directly from the relation.

While differentiating with respect to $x$ , remember that $y$ changes with $x$ . The chain rule gives $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$ . A term such as $x y$ needs the product rule: $\frac{d}{d x} (x y) = y + x \frac{d y}{d x}$ .

Collect the derivative terms

Differentiate every term, gather the terms containing $y^{'}$ on one side, then factor and divide. Substitute a point only after differentiating. First check that the point satisfies the original relation.

The resulting formula applies where its denominator is non-zero. A zero denominator can indicate a vertical tangent, which has no finite gradient. At singular points where both derivative coefficients vanish, investigate the original relation instead of dividing by zero.

2 x + 2 y \frac{d y}{d x} = 0

A common trap

Differentiate $y^{3}$ as $3 y^{2}$ .

Instead: Because differentiation is with respect to $x$ , the derivative is $3 y^{2} y^{'}$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Find the tangent to $x^{2} + y^{2} = 25$ at $(3, 4)$ .

Differentiate: $2 x + 2 y y^{'} = 0$ .
Solve: $y^{'} = - x / y$ , so the gradient is $- 3 / 4$ .
Use the point-gradient form with $(3, 4)$ .

y - 4 = - \frac{3}{4} (x - 3)

Example 2

Worked example

For $x y + y^{2} = 6$ , find $d y / d x$ and the normal at $(1, 2)$ .

Product and chain rules give $y + x y^{'} + 2 y y^{'} = 0$ .
Factor to obtain $y^{'} = - y / (x + 2 y)$ . At $(1, 2)$ the tangent gradient is $- 2 / 5$ .
The normal gradient is the negative reciprocal, $5 / 2$ .

y^{'} = - \frac{y}{x + 2 y}

; normal:

y - 2 = \frac{5}{2} (x - 1)

Check your understanding

Find $y^{'}$ for $x^{2} + 4 y^{2} = 16$ .

Show answer

$- x / (4 y)$

$2 x + 8 y y^{'} = 0$ , so $y^{'} = - x / (4 y)$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find $d y / d x$ for $x^{2} + 3 y^{2} = 12$ .
2 marks
Find $d y / d x$ for $x^{3} + y^{3} = 9$ .
2 marks
Find the tangent to $x^{2} + y^{2} = 10$ at $(1, 3)$ .
2 marks
Differentiate $x^{2} + x y + y^{2} = 7$ implicitly.
2 marks
For $x^{2} + y^{2} = 9$ , identify points where the tangent is vertical.
2 marks
Find the gradient of $sin y = x$ at $(0, 0)$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $d y / d x$ for $x^{2} + 3 y^{2} = 12$ .
2 marks
$2 x + 6 y y^{'} = 0$ .
Divide by $6 y$ , where $y \neq 0$ .
$- x / (3 y)$
Find $d y / d x$ for $x^{3} + y^{3} = 9$ .
2 marks
$3 x^{2} + 3 y^{2} y^{'} = 0$ .
$y^{'} = - x^{2} / y^{2}$ , for $y \neq 0$ .
$- x^{2} / y^{2}$
Find the tangent to $x^{2} + y^{2} = 10$ at $(1, 3)$ .
2 marks
The point satisfies $1 + 9 = 10$ .
$y^{'} = - x / y = - 1 / 3$ at the point.
$y - 3 = - \frac{1}{3} (x - 1)$
Differentiate $x^{2} + x y + y^{2} = 7$ implicitly.
2 marks
$2 x + y + x y^{'} + 2 y y^{'} = 0$ .
Collect $y^{'} (x + 2 y) = - (2 x + y)$ .
$y^{'} = - (2 x + y) / (x + 2 y)$
For $x^{2} + y^{2} = 9$ , identify points where the tangent is vertical.
2 marks
$y^{'} = - x / y$ has a zero denominator at $y = 0$ .
The circle then gives $x = \pm 3$ ; these are its left and right endpoints.
$(3, 0)$ and $(- 3, 0)$
Find the gradient of $sin y = x$ at $(0, 0)$ .
2 marks
$cos y y^{'} = 1$ .
At $(0, 0)$ , $cos 0 = 1$ , so $y^{'} = 1$ .
$1$

Checking saved progress…

Continue learning

Related-rate problems

Next lesson

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.1–6.2, with the approved school-level rates, differential-equation and motion applications.

Official curriculum source ↗

Back to Rates of change and differential equations