**Maintainer:**admin

*1*Chain rule¶

**Theorem:** Let $I, J$ be intervals in $\mathbb{R}$, $g : I \to \mathbb{R}$ and $f : J \to \mathbb{R}$ such that $f(J) \subseteq I$ and let $c \in J$. If $f$ is differentiable at $c$ and $g$ is differentiable at $f(c)$ then $g \circ f$ is differentiable at $c$ and $(g \circ f)'(c) = g'(f(c)) \cdot f'(c)$.

**Proof:** Since $f$ is differentiable at $c$, there exists a function $\phi$ continuous at $c$ such that $f(x) - f(c) = \phi(x)(x-c)$ for all $x \in J$. Similarly, there exists a function $\psi$ continuous at $f(c)$ such that for all $y \in I$, if $f(c) = d$ then $g(y) - g(d) = \psi(y)(y-d)$. Let $y = f(x)$. Since $f(J) \subseteq I$, $\psi(y)(y-d) = \psi(f(x))(f(x)-f(c)) = \psi(f(x))\phi(x)(x-c)$, i.e. $g(f(x)) - g(f(c)) = \psi(f(x))\phi(x)(x-c)$. Note that $\psi(f(x))\phi(x)$ is continuous at $c$. Thus by Carathéodory's Theorem, $g \circ f$ is differentiable at $c$. Furthermore, $(g \circ f)'(c) = \psi(f(c))\phi(c)$. As $\psi(f(c))$ is $g'(f(c))$ and $\phi(c)$ is $f'(c)$, we are done. $\blacksquare$

*2*Derivatives of inverse functions¶

If $y = f^{-1}(x)$, what is $\frac{dy}{dx}$?

**Theorem:** Let $I$ be an interval in $\mathbb{R}$ and $f : I \to \mathbb{R}$ be strictly monotone and continuous on $I$. Let $J = f(I)$. Let $f^{-1}: J \to \mathbb{R}$ be the strictly monotone and continuous inverse function. If $f$ is differentiable at $c \in I$ and $f'(c) \neq 0$ then $f^{-1}$ is differentiable at $f(c) = d$ and $(f^{-1})'(f(c)) = \frac{1}{f'(c)} = \frac{1}{f'(f^{-1}(d))}$.

**Proof:** By Carathéodory's Theorem there exists a function $\phi$ on $I$ continuous at $c$ such that $f(x) - f(c) = \phi(x)(x-c)$ and $f'(c) = \phi(c)$. By the hypotheses, $f'(c) \neq 0$ thus there exists some $\delta > 0$ such that $\phi(x) \neq 0$ for all $x \in I$ such that $|x-c|<\delta$ (i.e. $\phi$ is nonzero in some neighborhood of $c$). Let $V_\delta(c)$ be such a neighborhood. Let $U = f(V_\delta(c))$. Thus $f(f^{-1}(x)) = x$ for all $x \in U$. Hence:

$$\begin{align*} x-d &= f(f^{-1}(x)) - f(f^{-1}(d)) \\ &= \phi(f^{-1}(x))(f^{-1}(x) - c) \\ &= \phi(f^{-1}(x))(f^{-1}(x) - f^{-1}(d)) \\ \end{align*}$$

As $\phi(f^{-1}(x)) \neq 0$, it follows that $\displaystyle f^{-1}(x) - f^{-1}(d) = \frac{1}{\phi(f^{-1}(x))}(x-d)$ $\blacksquare$

*3*Relative extrema¶

**Definition:** Let $f : I \to \mathbb{R}$. $f$ is said to have a relative maximum at $c$ if there exists a neighborhood $V$ of $c$ where $f(x) \leq f(c)$ for all $x \in V$. The definition of relative minima is similar.

**Theorem:** Let $c$ be an interior point of the interval $I = [a,b]$, at which $f : I \to \mathbb{R}$ has a relative extremum. If $f$ is differentiable at $c$ then $f'(c) = 0$.

**Proof:** Suppose $f$ has a relative maximum at $c$. If $f'(c) > 0$ then there exists some neighborhood $V$ of $c$ on which $\displaystyle \frac{f(x)-f(c)}{x-c} > 0$. Let $x$ be an element of $V$ such that $x - c > 0$. Then $\displaystyle (x-c)\frac{f(x)-f(c)}{x-c} > 0 \implies f(x) > f(c)$. Thus $f$ doesn't have a relative maximum at $c$, thus $f'(c) \leq 0$. We can repeat this argument for $f'(c) < 0$, which shows $f'(c) \geq 0$. Thus $f'(c) = 0$. The proof is similar for relative minima. $\blacksquare$

**Corollary:** Let $I$ be an interval, $f : I \to \mathbb{R}$ continuous and suppose $f$ has a relative extremum at an interior point $c \in I$. Then either $f'(c) = 0$ or $f$ is not differentiable at $c$.