IB Analysis and Approaches HL · Differential calculus · Section 14 of 14

Implicit differentiation

3 worked examples · By Álvaro López

Key idea

Some curves have equations that mix $x$ and $y$ and cannot (or should not) be solved for $y$ , such as $x^{2} + x y + y^{2} = 3$ . Implicit differentiation lets you find $\frac{d y}{d x}$ by differentiating the equation as it stands. With it you obtain gradients of tangents, horizontal tangents and second derivatives. This is Higher Level content and is examined mainly in Paper 1.

Implicit function

A curve is given in implicit form when it is described by an equation $F (x, y) = 0$ instead of $y = f (x)$ . A circle, for example $x^{2} + y^{2} = 25$ , is not the graph of a single function of $x$ . Even so, near each point of the curve, $y$ behaves like a differentiable function of $x$ , and that is enough to differentiate.

The rule

Differentiate both sides with respect to $x$ , remembering that $y$ depends on $x$ . Every term containing $y$ is differentiated using the chain rule: a factor $\frac{d y}{d x}$ (or $y^{'}$ ) appears.

The reason: $y^{2}$ is a composite function, $(y (x))^{2}$ , and its derivative is $2 y \cdot y^{'}$ , just as the derivative of $(something)^{2}$ is $2 \cdot (something) \cdot (something)^{'}$ . And $x y$ is a product of two functions of $x$ , so the product rule is used.

Second implicit derivative

To find $y^{''}$ , differentiate the expression for $y^{'}$ (or the already differentiated equation) again and substitute $y^{'}$ by its expression. If $y^{''}$ is asked at a point where $y^{'} = 0$ , the calculation is simplified: the terms with $y^{'}$ disappear.

Worked examples

Example 1 Basic

Gradient on an ellipse

The curve $x^{2} + 4 y^{2} = 20$ passes through the point $(2, 2)$ .

a) Find $\frac{d y}{d x}$ in terms of $x$ and $y$ .

b) Find the gradient of the tangent at $(2, 2)$ .

Solution

a) Differentiate with respect to $x$ : $2 x + 8 y \frac{d y}{d x} = 0$ . Solve: $\frac{d y}{d x} = - \frac{2 x}{8 y} = - \frac{x}{4 y} .$

b) At $(2, 2)$ : $\frac{d y}{d x} = - \frac{2}{8} = - \frac{1}{4}$ .

Example 2 Medium

Tangent and normal

The curve $x^{2} y + y^{3} = 10$ passes through the point $P (1, 2)$ .

a) Find $\frac{d y}{d x}$ in terms of $x$ and $y$ .

b) Find the gradient of the tangent at $P$ .

c) Find the equations of the tangent and the normal at $P$ , in the form $y = m x + n$ .

Solution

a) Differentiate with respect to $x$ , using the product rule on the first term: $2 x y + x^{2} \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x} = 0.$ Group the terms: $\frac{d y}{d x} (x^{2} + 3 y^{2}) = - 2 x y$ , so $\frac{d y}{d x} = - \frac{2 x y}{x ^{2} + 3 y ^{2}} .$

b) At $(1, 2)$ : $\frac{d y}{d x} = - \frac{4}{1 + 12} = - \frac{4}{13}$ .

c) Tangent: $y - 2 = - \frac{4}{13} (x - 1)$ , that is, $y = - \frac{4}{13} x + \frac{30}{13}$ . The normal has gradient $\frac{13}{4}$ : $y - 2 = \frac{13}{4} (x - 1)$ , that is, $y = \frac{13}{4} x - \frac{5}{4}$ .

Example 3 Hard

Horizontal tangents and the second derivative

Let the curve be $x^{2} + x y + y^{2} = 3$ .

a) Find $\frac{d y}{d x}$ .

b) Find the points on the curve at which the tangent is horizontal.

c) Find $\frac{d ^{2} y}{d x ^{2}}$ at each of those points (in order of increasing $x$ -coordinate) and decide whether it is a maximum or a minimum of $y$ .

Solution

a) Differentiate: $2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$ , that is, $(x + 2 y) \frac{d y}{d x} = - (2 x + y)$ : $\frac{d y}{d x} = - \frac{2 x + y}{x + 2 y} .$

b) The tangent is horizontal if $2 x + y = 0$ , that is $y = - 2 x$ . Substitute into the equation of the curve: $x^{2} + x (- 2 x) + 4 x^{2} = 3 x^{2} = 3 \Rightarrow x = \pm 1$ . The points are $(- 1, 2)$ and $(1, - 2)$ (in both, $x + 2 y \neq = 0$ ).

c) Differentiate the equation $2 x + y + (x + 2 y) y^{'} = 0$ once more with respect to $x$ : $2 + y^{'} + (1 + 2 y^{'}) y^{'} + (x + 2 y) y^{''} = 0.$ At the points where $y^{'} = 0$ this gives $2 + (x + 2 y) y^{''} = 0$ , so $y^{''} = - \frac{2}{x + 2 y}$ .

At $(- 1, 2)$ : $x + 2 y = 3$ , $y^{''} = - \frac{2}{3} < 0$ : a maximum of $y$ .
At $(1, - 2)$ : $x + 2 y = - 3$ , $y^{''} = \frac{2}{3} > 0$ : a minimum of $y$ .

In the exam: write each derivative clearly and justify that the point lies on the curve. For $y^{''}$ , differentiate the expression already grouped and substitute $y^{'}$ , or use that $y^{'} = 0$ at those points, as here. Check that the points with a horizontal tangent also satisfy the original equation.

Common mistakes

Summary

Differentiate the equation with respect to $x$ ; each term with $y$ carries the factor $y^{'}$ (chain rule), and $x y$ uses the product rule.
Group the terms in $y^{'}$ , take out the common factor and solve.
Gradient at a point: substitute $x$ and $y$ . Tangent: $y - y_{0} = m (x - x_{0})$ ; normal: gradient $- \frac{1}{m}$ .
Horizontal tangent: numerator of $y^{'}$ equal to $0$ ; vertical: denominator equal to $0$ . The point must lie on the curve.
$y^{''}$ : differentiate again and substitute $y^{'}$ .

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