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Derivative rules

Implicit Differentiation: Steps, Formula and Examples

Implicit differentiation finds dy/dx when y is not solved for. The key rule, a four-step method, worked examples with tangent lines and common mistakes.

Derivative rules Updated 2 min read

At a glance

\frac{d}{dx}\,f(y) = f'(y)\,\frac{dy}{dx}
Applies to
Equations relating x and y where y is not isolated, such as x^2 + y^2 = 25
Key idea
Treat y as a function of x, so every y term gets a factor of \frac{dy}{dx} by the chain rule
Result
A formula for \frac{dy}{dx} that usually contains both x and y
Needs
Chain rule, and the product rule for terms like xy
Used in
Tangent lines to curves, related rates, derivatives of inverse functions
On this page
  1. The key rule
  2. The four steps
  3. Worked examples
  4. When to use implicit differentiation
  5. Common mistakes

Implicit differentiation finds \frac{dy}{dx} from an equation that mixes x and y, without first solving for y. You differentiate both sides with respect to x, remember that y depends on x, and then solve for \frac{dy}{dx}. It is the chain rule applied to y.

The key rule

Since y is a function of x, any expression in y is a composition. By the chain rule:

\frac{d}{dx}\,f(y) = f'(y)\,\frac{dy}{dx}
Term Derivative with respect to x
x^3 3x^2
y^3 3y^2\,\frac{dy}{dx}
\sin y \cos y\,\frac{dy}{dx}
e^y e^y\,\frac{dy}{dx}
xy y + x\,\frac{dy}{dx} (product rule)

The four steps

  1. Differentiate both sides of the equation with respect to x.
  2. Multiply by \frac{dy}{dx} every time you differentiate a y term.
  3. Move all \frac{dy}{dx} terms to one side and everything else to the other.
  4. Factor out \frac{dy}{dx} and divide.

Worked examples

Example 1: a circle. Differentiate x^2 + y^2 = 25:

2x + 2y\,\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{x}{y}

At the point (3, 4) the slope is -\frac{3}{4}, so the tangent line is y - 4 = -\frac{3}{4}(x - 3), or y = -\frac{3}{4}x + \frac{25}{4}.

Circle x² + y² = 25 with its tangent line at (3, 4)
Implicit differentiation gives \frac{dy}{dx} = -\frac{x}{y}, so the slope at (3, 4) is -\frac{3}{4}.

Example 2: a product term. Differentiate x^3 + y^3 = 6xy, using the product rule on 6xy:

3x^2 + 3y^2\,\frac{dy}{dx} = 6y + 6x\,\frac{dy}{dx}

Collect and factor:

\frac{dy}{dx}\left(3y^2 - 6x\right) = 6y - 3x^2 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}

The point (3, 3) is on the curve, since 27 + 27 = 54 = 6 \cdot 9. There the slope is \frac{6 - 9}{9 - 6} = -1.

Example 3: recovering the derivative of \ln x. If y = \ln x then e^y = x. Differentiate:

e^y\,\frac{dy}{dx} = 1 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}

This is the idea behind the inverse function rule.

Example 4: a trig equation. Differentiate \sin y = x^2:

\cos y\,\frac{dy}{dx} = 2x \quad\Rightarrow\quad \frac{dy}{dx} = \frac{2x}{\cos y}

When to use implicit differentiation

  • The equation is hard or impossible to solve for y (like x^3 + y^3 = 6xy).
  • The curve is not a single function, like a circle or an ellipse, and you want slopes on every part.
  • You need derivatives of inverse functions such as \ln x, \arcsin x or \arctan x.
  • In related rates problems, where you differentiate with respect to time t and every quantity gets its own rate.

Common mistakes

  1. Forgetting \frac{dy}{dx} on y terms. \frac{d}{dx}\,y^2 is 2y\,\frac{dy}{dx}, not 2y.
  2. Skipping the product rule. \frac{d}{dx}\,xy is y + x\,\frac{dy}{dx}, not \frac{dy}{dx} or 1 \cdot \frac{dy}{dx}.
  3. Differentiating only one side. Both sides change, including a constant side, whose derivative is 0.
  4. Plugging in a point that is not on the curve. Check the point satisfies the original equation before using it.

For a longer walkthrough with more practice problems, read the full guide to implicit differentiation.

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Frequently asked questions

What is implicit differentiation?
It is a way to find \frac{dy}{dx} from an equation in x and y without solving for y. You differentiate both sides with respect to x and use the chain rule on every y term.

Why do you multiply by dy/dx?
Because y is a function of x. By the chain rule, \frac{d}{dx}\,y^2 = 2y \cdot \frac{dy}{dx}, just as \frac{d}{dx}\,(g(x))^2 = 2g(x)\,g'(x).

When do you use implicit differentiation?
When y is hard or impossible to isolate, as in x^3 + y^3 = 6xy, or when the curve is not the graph of a single function, like a circle.

Why does the answer have y in it?
The curve can pass through several points with the same x and different slopes. The y tells you which point you mean.

Written by the PracticeCalculus team

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Related rules

Chain rule \frac{d}{dx}\,f(g(x)) = f'(g(x))\,g'(x) The chain rule: the derivative of f(g(x)) is f'(g(x))·g'(x). The formula in both notations, a table of common forms, worked examples and mistakes. Product rule \frac{d}{dx}\,[f(x)\,g(x)] = f'(x)\,g(x) + f(x)\,g'(x) The product rule: the derivative of f·g is f'g + fg'. The formula, worked examples, the three-factor version, a short proof and common mistakes. Inverse function rule (f^{-1})'(x) = \frac{1}{f'\big(f^{-1}(x)\big)} The inverse function rule: the derivative of f⁻¹ at x is 1/f'(f⁻¹(x)). The formula, why slopes are reciprocals, worked examples and arcsin and ln.