Implicit differentiation | Year 12

01 / LEARN

Understand the idea.

Treat y as a function of x

An implicit relation can describe several local branches without isolating $y$ . Differentiate the equation with respect to $x$ while treating $y$ as a function of $x$ . Thus $\frac{d}{d x} (y^{2}) = 2 y y^{'}$ by the chain rule.

A mixed term needs the product rule: $\frac{d}{d x} (x y) = y + x y^{'}$ . Write every derivative before collecting terms. Substituting a point too soon would turn changing coordinates into constants and destroy the relation you need to differentiate.

Collect the derivative terms

Gather the terms containing $y^{'}$ , factor it out, and divide by its coefficient only where that coefficient is nonzero. Check that a stated point lies on the curve before substituting it into the gradient formula.

A tangent uses the local gradient and point-gradient form. A normal has the negative reciprocal gradient when the tangent gradient is finite and nonzero. A zero denominator or a singular point needs separate investigation; division alone cannot classify it.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

A circle centred at O with radius 5, point P at (3,4), and the tangent through P. The tangent is perpendicular to OP. Both axes use equal physical scales.

Worked solution

1
Check the point and differentiate.
$3^{2} + 4^{2} = 25, 2 x + 2 y y^{'} = 0$
The point lies on the circle, and differentiating y squared requires a chain factor.
2
Find the tangent gradient.
$y^{'} = - \frac{x}{y}, m = - \frac{3}{4}$
At the stated point, y is nonzero, so division is valid.
3
Write the tangent through the point.
$y - 4 = - \frac{3}{4} (x - 3)$
Point-gradient form gives the line with gradient minus three quarters through (3,4).
4
Check the geometry.
$\frac{4}{3} (- \frac{3}{4}) = - 1$
The radius to (3,4) has gradient four thirds, confirming perpendicularity with the tangent.

Answer

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

EXAMPLE 02

Worked example

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

Worked solution

1
Check the point and differentiate the relation.
$1 \cdot 2 + 2^{2} = 6, y + x y^{'} + 2 y y^{'} = 0$
The point is valid; xy needs the product rule and y squared needs the chain rule.
2
Collect the derivative terms.
$(x + 2 y) y^{'} = - y, y^{'} = - \frac{y}{x + 2 y}$
Factoring isolates the derivative where x plus 2y is nonzero.
3
Find the normal gradient.
$m_{\tan} = - \frac{2}{1 + 4} = - \frac{2}{5}, m_{normal} = \frac{5}{2}$
The normal is perpendicular to the tangent, so their gradients have product minus one.
4
Write the normal line.
$y - 2 = \frac{5}{2} (x - 1)$
The normal passes through the same point as the tangent.

Answer

$y^{'} = - \frac{y}{x + 2 y}$ ; normal: $y - 2 = \frac{5}{2} (x - 1)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$- x / (4 y)$

Implicit differentiation gives $2 x + 8 y y^{'} = 0$ . Dividing by $8 y$ yields $y^{'} = - \frac{x}{4 y}$ where $y \neq 0$ ; the points with $y = 0$ do not have a finite gradient from this formula.

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6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find $\frac{d y}{d x}$ for $x^{2} + 3 y^{2} = 12$ .
2 marks
Find $\frac{d y}{d x}$ 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 $\frac{d y}{d x}$ for $x^{2} + 3 y^{2} = 12$ .
2 marks
$2 x + 6 y y^{'} = 0$ .
Divide by $6 y$ , where $y \neq 0$ .
$- \frac{x}{3 y}$
Find $\frac{d y}{d x}$ for $x^{3} + y^{3} = 9$ .
2 marks
$3 x^{2} + 3 y^{2} y^{'} = 0$ .
$y^{'} = - \frac{x^{2}}{y^{2}}$ , for $y \neq 0$ .
$- \frac{x^{2}}{y^{2}}$
Find the tangent to $x^{2} + y^{2} = 10$ at $(1, 3)$ .
2 marks
The point satisfies $1 + 9 = 10$ .
$y^{'} = - \frac{x}{y} = - \frac{1}{3}$ 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 $y^{'} (x + 2 y) = - (2 x + y)$ .
$y^{'} = - \frac{2 x + y}{x + 2 y}$
For $x^{2} + y^{2} = 9$ , identify points where the tangent is vertical.
2 marks
$y^{'} = - \frac{x}{y}$ 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 $y^{'} = 1$ .
$1$

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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 ↗

In this lesson

Understand the idea.

Treat y as a function of x

Collect the derivative terms

See the method in action.

Worked example

Check the point and differentiate.

Find the tangent gradient.

Write the tangent through the point.

Check the geometry.

Worked example

Check the point and differentiate the relation.

Collect the derivative terms.

Find the normal gradient.

Write the normal line.

Try it yourself.

Check your understanding

Keep practising