Implicit Differentiation

When to Use

Use implicit differentiation when:

• $y$ is not isolated (e.g., $x^2 + y^2 = 25$)
• The equation is difficult or impossible to solve for $y$
• You need $\frac{dy}{dx}$ without explicitly solving for $y$

The Method

1. Differentiate both sides with respect to $x$
2. Apply the chain rule to terms containing $y$ (multiply by $\frac{dy}{dx}$)
3. Solve for $\frac{dy}{dx}$

Key Principle

When differentiating $y$ with respect to $x$:

$\frac{d}{dx}[y] = \frac{dy}{dx}$

$\frac{d}{dx}[y^n] = n \cdot y^{n-1} \cdot \frac{dy}{dx}$

$\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}$

Examples

Example 1: Circle

Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$:

$\frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25]$

$2x + 2y \cdot \frac{dy}{dx} = 0$

$\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$

Example 2: Product of x and y

Find $\frac{dy}{dx}$ for $xy = 1$:

$\frac{d}{dx}[xy] = \frac{d}{dx}[1]$

$x \cdot \frac{dy}{dx} + y \cdot 1 = 0$

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

Example 3: More complex equation

Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$:

$\frac{d}{dx}[x^3 + y^3] = \frac{d}{dx}[6xy]$

$3x^2 + 3y^2 \cdot \frac{dy}{dx} = 6\left(x \cdot \frac{dy}{dx} + y \cdot 1\right)$

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

$3y^2 \cdot \frac{dy}{dx} - 6x \cdot \frac{dy}{dx} = 6y - 3x^2$

$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$

$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$

Example 4: With trig functions

Find $\frac{dy}{dx}$ for $\sin(xy) = x$:

$\cos(xy) \cdot \frac{d}{dx}[xy] = 1$

$\cos(xy) \cdot \left(x \cdot \frac{dy}{dx} + y\right) = 1$

$x \cos(xy) \cdot \frac{dy}{dx} + y \cos(xy) = 1$

$\frac{dy}{dx} = \frac{1 - y \cos(xy)}{x \cos(xy)}$

Example 5: Finding the tangent line

Find the tangent line to $x^2 + xy + y^2 = 7$ at $(1, 2)$:

First, find $\frac{dy}{dx}$:

$2x + x \cdot \frac{dy}{dx} + y + 2y \cdot \frac{dy}{dx} = 0$

$\frac{dy}{dx}(x + 2y) = -2x - y$

$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

At $(1, 2)$:

$\frac{dy}{dx} = -\frac{2(1) + 2}{1 + 2(2)} = -\frac{4}{5}$

Tangent line: $y - 2 = -\frac{4}{5}(x - 1)$, or $y = -\frac{4x}{5} + \frac{14}{5}$

Second Derivatives

To find $\frac{d^2y}{dx^2}$, differentiate $\frac{dy}{dx}$ implicitly again:

Example

For $x^2 + y^2 = 25$, we found $\frac{dy}{dx} = -\frac{x}{y}$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x}{y}\right]$

$= -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2}$

$= -\frac{y - x \cdot \left(-\frac{x}{y}\right)}{y^2}$

$= -\frac{y + \frac{x^2}{y}}{y^2}$

$= -\frac{y^2 + x^2}{y^3}$

$= -\frac{25}{y^3} \quad \text{(since } x^2 + y^2 = 25\text{)}$