# 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.

Source: https://practicecalculus.com/rules/implicit-differentiation
Updated: 2026-11-03

$$ \frac{d}{dx}\,f(y) = f'(y)\,\frac{dy}{dx} $$

## At a glance

- **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

## Key takeaways

- Differentiate both sides with respect to $x$, treating $y$ as a function of $x$.
- Every time you differentiate a term with $y$, multiply by $\frac{dy}{dx}$.
- Collect the $\frac{dy}{dx}$ terms on one side and factor to solve.
- The answer usually involves both $x$ and $y$. Plug in a point on the curve to get a slope.

**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](/rules/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 $y = -\frac{3}{4}x + \frac{25}{4}$.

```graph
title: 'Circle x² + y² = 25 with its tangent line at (3, 4)'
caption: 'Implicit differentiation gives $\frac{dy}{dx} = -\frac{x}{y}$, so the slope at $(3, 4)$ is $-\frac{3}{4}$.'
x: [-8, 10]
y: [-6, 7]
functions:
  - expr: sqrt(25 - x^2)
    label: upper half
    domain: [-5, 5]
  - expr: -sqrt(25 - x^2)
    label: lower half
    domain: [-5, 5]
  - expr: -0.75x + 6.25
    label: tangent at (3, 4)
    dashed: true
points:
  - [3, 4, '(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](/rules/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](/blog/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](/blog/implicit-differentiation).

## 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.