$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

Using Leibniz notation: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Combine chain rule with other differentiation rules:

• $(e^{f(x)})' = e^{f(x)} \cdot f'(x)$
• $(\sin(f(x)))' = \cos(f(x)) \cdot f'(x)$
• $(\ln(f(x)))' = \frac{f'(x)}{f(x)}$

1. Differentiate both sides with respect to x
2. Apply chain rule to y-terms: $\frac{d}{dx}(y^n) = ny^{n-1} \cdot \frac{dy}{dx}$
3. Collect all $\frac{dy}{dx}$ terms on one side
4. Solve for $\frac{dy}{dx}$

Example: $x^2 + y^2 = 25$ $2x + 2y \cdot \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{x}{y}$

First find $\frac{dy}{dx}$ implicitly, then differentiate again with respect to x, applying chain rule and product rule as needed.

If g = f-1, then: $g'(x) = \frac{1}{f'(g(x))}$

Alternatively: if f(a) = b and f'(a) exists and ≠ 0, then: $(f^{-1})'(b) = \frac{1}{f'(a)}$

Exponential: $\frac{d}{dx}(a^x) = a^x \ln a$

Logarithmic: $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$

$\frac{d}{dx}(\ln x) = \frac{1}{x}$

See inverse trigonometric functions and inverse functions in general above.

Higher-Order Derivatives (Second and Higher)

Second derivative: $f''(x) = \frac{d}{dx}[f'(x)] = \frac{d^2y}{dx^2}$

nth derivative: $f^{(n)}(x) = \frac{d^n}{dx^n}[f(x)]$

Notation: f''(x), y''', f(n)(x), or $\frac{d^ny}{dx^n}$