Implicit Differentiation

An equation in

x

and

y

, like

x^2 + y^2 = 100

, often isn't the graph of one function. The circle fails the vertical line test. Near most of its points, though, the curve is the graph of a differentiable function, and implicit differentiation finds its slope without solving for

y

Implicit differentiation

To find

\frac{dy}{dx}

from an equation in

x

and

y

Differentiate both sides with respect to

x

, treating

y

as a differentiable function of

x

. Every term with

y

needs the chain rule, so

\begin{aligned} \frac{d}{dx}\big[y^n\big] &= ny^{n - 1}\frac{dy}{dx}, \\[4pt] \frac{d}{dx}\big[\sin y\big] &= \cos y\,\frac{dy}{dx}, \end{aligned}

and a term like

xy

needs the product rule as well. Then collect the

\frac{dy}{dx}

terms on one side and solve for

\frac{dy}{dx}

. The result holds wherever its denominator isn't

0

When a problem gives

\frac{dy}{dx}

and asks you to show it, the last line must be that exact expression rather than an equivalent form.

The answer usually contains both

x

and

y

, so the slope at a point needs both coordinates.

Tangent lines

Example 1

Find

\frac{dy}{dx}

for the circle

x^2 + y^2 = 100

, and the tangent line at

(-6, 8)

Differentiate both sides:

\begin{aligned} 2x + 2y\frac{dy}{dx} &= 0, \\ \frac{dy}{dx} &= -\frac{x}{y} \quad (y \ne 0). \end{aligned}

(-6, 8)

the slope is

-\frac{-6}{8} = \frac{3}{4}

, and the tangent line is

y = 8 + \frac{3}{4}(x + 6)

The upper half of the circle is

y = \sqrt{100 - x^2}

, and differentiating it directly gives

\frac{-x}{\sqrt{100 - x^2}} = -\frac{x}{y}

, the same answer. The implicit version covers both halves at once: at

(-6, -8)

it gives slope

-\frac{3}{4}

The circle x² + y² = 100 and its tangent line at (−6, 8), with slope 3/4.

Example 2

The curve

x\cos y + y = 2

passes through

(2, 0)

. Find the tangent line there.

The term

x\cos y

is a product, and

\cos y

needs the chain rule. Differentiating gives

\cos y - x\sin y\,\frac{dy}{dx} + \frac{dy}{dx} = 0

. Collect the

\frac{dy}{dx}

terms:

\begin{aligned} (1 - x\sin y)\frac{dy}{dx} &= -\cos y, \\ \frac{dy}{dx} &= \frac{-\cos y}{1 - x\sin y}. \end{aligned}

(2, 0)

\frac{dy}{dx} = \frac{-1}{1 - 0} = -1

, and the tangent line is

y = -(x - 2)

, or

y = 2 - x

Part of the curve x cos y + y = 2 and its tangent line y = 2 − x at (2, 0).

Horizontal and vertical tangents

When

\frac{dy}{dx}

is a fraction, the tangent is horizontal at a point of the curve where the numerator is

0

and the denominator isn't. It's vertical where the denominator is

0

and the numerator isn't. Where both are

0

, the formula decides nothing, as at the origin in Example 4.

Example 3

Find the points on

x^2 - xy + y^2 = 3

where the tangent line is horizontal or vertical.

Differentiate, using the product rule on

xy

\begin{aligned} 2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} &= 0, \\ (2y - x)\frac{dy}{dx} &= y - 2x, \\ \frac{dy}{dx} &= \frac{y - 2x}{2y - x}. \end{aligned}

Horizontal:

y = 2x

. Substituting into the curve gives

x^2 - 2x^2 + 4x^2 = 3

, so

x = \pm 1

. The points are

(1, 2)

and

(-1, -2)

, where the denominator is

3

and

-3

Vertical:

x = 2y

. Substituting gives

4y^2 - 2y^2 + y^2 = 3

, so

y = \pm 1

. The points are

(2, 1)

and

(-2, -1)

, where the numerator is

-3

and

3

The tilted ellipse x² − xy + y² = 3. Horizontal tangents at (1, 2) and (−1, −2); vertical tangents at (2, 1) and (−2, −1).

Example 4

The curve

x^4 + y^4 = 4xy

is shown below. Find

\frac{dy}{dx}

, and find the point in the first quadrant where the tangent line is horizontal.

\begin{aligned} 4x^3 + 4y^3\frac{dy}{dx} &= 4y + 4x\frac{dy}{dx}, \\ \frac{dy}{dx} &= \frac{y - x^3}{y^3 - x}. \end{aligned}

The numerator is

0

when

y = x^3

. Substituting into the curve,

\begin{aligned} x^4 + x^{12} &= 4x^4, \\ x^4\left(x^8 - 3\right) &= 0. \end{aligned}

x = 0

the numerator and the denominator are both

0

, so the formula gives no slope at the origin, where the two loops meet. The other root in the first quadrant is

x = 3^{1/8} \approx 1.15

, with

y = x^3 = 3^{3/8} \approx 1.51

. There the denominator is

3^{9/8} - 3^{1/8} = 2 \cdot 3^{1/8} \ne 0

, so the tangent is horizontal.

The curve x⁴ + y⁴ = 4xy, two loops that meet at the origin. The horizontal tangent in the first quadrant is at (3^(1/8), 3^(3/8)); by symmetry across y = x, a vertical tangent is at (3^(3/8), 3^(1/8)).