Implicit Differentiation

Quick answer

An equation such as x² + y² = 25 defines y as a function of x implicitly, without a formula y = f(x). To find dy/dx, differentiate both sides with respect to x, treating y as an unknown function of x: by the chain rule, every term containing y picks up a factor dy/dx. Then solve the resulting equation for dy/dx. The slope usually depends on both x and y, which is what lets one curve have different slopes at points with the same x.

What you'll learn

Differentiate an equation implicitly with respect to x
Solve for dy/dx and evaluate it at a point
Find tangent lines to implicitly defined curves
Find a second derivative implicitly

When y is not solved for

The circle $x^2 + y^2 = 25$ is not the graph of a single function: each $x$ between $-5$ and $5$ has two $y$ values. Near any one point, though, the circle is the graph of some function — the top half is $y = \sqrt{25 - x^2}$ — and it has a tangent line.

tangent: y = −¾x + 25/4

A circle and its tangent line at (3, 4)

Solving for $y$ works here but gets ugly fast, and for curves such as $x^3 + y^3 = 6xy$ it is hopeless. Implicit differentiation finds the slope without ever solving for $y$ .

The method

Differentiate both sides of the equation with respect to $x$ .
Treat $y$ as a function of $x$ : every term with $y$ gets a factor of $\tfrac{dy}{dx}$ from the chain rule.
Collect the $\tfrac{dy}{dx}$ terms on one side and solve.

For the circle:

\frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx}\,25 \quad\Longrightarrow\quad 2x + 2y\frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{x}{y}

Why every y term gets a dy/dx

$y$ stands for a function of $x$ , even if we never write its formula. So $y^2$ is really $\big(y(x)\big)^2$ — a composite — and the chain rule gives

\frac{d}{dx}\,y^2 = 2y \cdot \frac{dy}{dx}

Compare $\tfrac{d}{dx}\,x^2 = 2x \cdot \tfrac{dx}{dx} = 2x$ : the factor is there too, but it equals $1$ . Implicit differentiation is the chain rule applied to a function whose formula is unknown. The factor $\tfrac{dy}{dx}$ is the inner derivative that the unknown function contributes.

Second derivatives

Differentiate $\tfrac{dy}{dx}$ again, remembering that $y$ inside it is still a function of $x$ . For the circle, the quotient rule gives

\frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3} = -\frac{25}{y^3}

The last step uses the original equation, $x^2 + y^2 = 25$ . On the top half ( $y > 0$ ) the second derivative is negative, so the arc bends downward.

Worked examples

Common mistakes

Practice problems

Find $\tfrac{dy}{dx}$ for $x^2 + 4y^2 = 16$ .

Answer

$\tfrac{dy}{dx} = -\tfrac{x}{4y}$

Full solution

$2x + 8y\tfrac{dy}{dx} = 0$ , so $\tfrac{dy}{dx} = -\tfrac{2x}{8y} = -\tfrac{x}{4y}$ .
Find $\tfrac{dy}{dx}$ for $xy = 12$ .

Answer

$\tfrac{dy}{dx} = -\tfrac{y}{x}$

Full solution

Product rule: $y + x\tfrac{dy}{dx} = 0$ , so $\tfrac{dy}{dx} = -\tfrac{y}{x}$ .
Find the tangent line to $x^2 + y^2 = 169$ at $(5, -12)$ .

Answer

$y = \tfrac{5}{12}x - \tfrac{169}{12}$

Full solution

$\tfrac{dy}{dx} = -\tfrac{x}{y} = -\tfrac{5}{-12} = \tfrac{5}{12}$ . Then $y + 12 = \tfrac{5}{12}(x - 5)$ , so $y = \tfrac{5}{12}x - \tfrac{25}{12} - 12 = \tfrac{5}{12}x - \tfrac{169}{12}$ .
Find $\tfrac{dy}{dx}$ for $y^3 + y = x^2$ .

Answer

$\tfrac{dy}{dx} = \tfrac{2x}{3y^2 + 1}$

Full solution

$3y^2\tfrac{dy}{dx} + \tfrac{dy}{dx} = 2x$ , so $(3y^2 + 1)\tfrac{dy}{dx} = 2x$ .
Find $\tfrac{dy}{dx}$ for $x^2 y + y^2 = 7$ .

Answer

$\tfrac{dy}{dx} = -\tfrac{2xy}{x^2 + 2y}$

Full solution

Product rule on $x^2 y$ : $2xy + x^2\tfrac{dy}{dx} + 2y\tfrac{dy}{dx} = 0$ . So $(x^2 + 2y)\tfrac{dy}{dx} = -2xy$ .
Find $\tfrac{dy}{dx}$ for $e^y = x + y$ .

Answer

$\tfrac{dy}{dx} = \tfrac{1}{e^y - 1}$

Full solution

$e^y\tfrac{dy}{dx} = 1 + \tfrac{dy}{dx}$ , so $(e^y - 1)\tfrac{dy}{dx} = 1$ .
At which points of $x^2 + y^2 = 25$ is the tangent line horizontal? Vertical?

Answer

Horizontal at $(0, \pm 5)$ ; vertical at $(\pm 5, 0)$ .

Full solution

$\tfrac{dy}{dx} = -\tfrac{x}{y}$ is $0$ when $x = 0$ (with $y = \pm 5$ ), and undefined when $y = 0$ (with $x = \pm 5$ ), where the tangent is vertical.
Find $\tfrac{d^2y}{dx^2}$ for $xy = 1$ , in terms of $x$ and $y$ .

Answer

$\tfrac{2y}{x^2}$

Full solution

From problem 2’s method, $\tfrac{dy}{dx} = -\tfrac{y}{x}$ . Quotient rule: $\tfrac{d^2y}{dx^2} = -\tfrac{x\tfrac{dy}{dx} - y}{x^2} = -\tfrac{-y - y}{x^2} = \tfrac{2y}{x^2}$ .
Find the slope of $x^3 + y^3 = 6xy$ at $\left(\tfrac{4}{3}, \tfrac{8}{3}\right)$ , a point on the curve.

Answer

$\tfrac{4}{5}$

Full solution

The point is on the curve: $x^3 + y^3 = \tfrac{64}{27} + \tfrac{512}{27} = \tfrac{64}{3}$ and $6xy = 6 \cdot \tfrac{32}{9} = \tfrac{64}{3}$ .

Use $\tfrac{dy}{dx} = \tfrac{2y - x^2}{y^2 - 2x}$ from Example 2. The top is $\tfrac{48}{9} - \tfrac{16}{9} = \tfrac{32}{9}$ and the bottom is $\tfrac{64}{9} - \tfrac{24}{9} = \tfrac{40}{9}$ .

The slope is $\tfrac{32}{40} = \tfrac{4}{5}$ .
A student differentiates $x^2 + y^2 = 25$ to get $2x + 2y = 0$ . What went wrong?

Hint

$y$ is a function of $x$ .

Answer

The $\tfrac{dy}{dx}$ from the chain rule is missing: $2x + 2y\tfrac{dy}{dx} = 0$ .

Full solution

$\tfrac{d}{dx}\,y^2 = 2y\tfrac{dy}{dx}$ , because $y$ depends on $x$ .

The student’s equation, $2x + 2y = 0$ , is a line, not a statement about slopes.

Frequently asked questions

What is implicit differentiation?

Differentiating both sides of an equation in x and y with respect to x, treating y as a function of x, then solving for dy/dx.

Why does the derivative of y² have a dy/dx in it?

Because y is a function of x, y² is a composite, and the chain rule gives 2y · dy/dx.

How do I differentiate xy implicitly?

With the product rule: d/dx (xy) = 1 · y + x · dy/dx.

Why does the answer contain both x and y?

An implicit curve can pass through several points with the same x — a circle does — and each has its own slope, so the slope needs y to say which point is meant.

When should I use implicit differentiation?

When solving for y is hard or impossible, or gives more than one branch, as with a circle or x³ + y³ = 6xy.

What to learn next

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