Differentiation: Composite, Implicit, and Inverse Functions

Notes

The chain rule

For a composite $f(g(x))$, take the derivative of the outside and multiply by the derivative of the inside:

• $\displaystyle \frac{d}{dx}(3x + 1)^5 = 5(3x + 1)^4 \cdot 3$.
• $\displaystyle \frac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x$.
• $\displaystyle \frac{d}{dx}e^{u} = e^u u'$ and $\displaystyle \frac{d}{dx}\ln u = \frac{u'}{u}$.
• Other bases: $\displaystyle \frac{d}{dx}a^x = a^x\ln a$ and $\displaystyle \frac{d}{dx}\log_a x = \frac{1}{x\ln a}$.

Implicit differentiation

When $y$ isn't solved for, differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Every $y$ term gets a factor $\displaystyle \frac{dy}{dx}$.

• Example: $\displaystyle x^2 + y^2 = 25 \Rightarrow 2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.
• Second derivatives: differentiate $\displaystyle \frac{dy}{dx}$ again, then substitute $\displaystyle \frac{dy}{dx}$ back in.

Inverse functions

If $g = f^{-1}$ and $f(a) = b$, then

• Inverse trig: $\displaystyle \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$, $\displaystyle \frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1 - x^2}}$, $\displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$.

Higher-order derivatives

$f''(x)$ is the derivative of $f'(x)$. If $s(t)$ is position, $s'(t)$ is velocity and $s''(t)$ is acceleration.