7. Derivatives

Lemma

Is $f$ at $x_0$ differentiable, then $f$ is continuous at that point.

One-sided derivatives: $\underset{\begin{array}{c}h\to 0\\ h>0\end{array}}{\lim }\frac{f(a+h)-f(a)}{h}$

Higher derivatives: $C^n(D)$ is the set of all $n$-times differentiable function on $D$. $C^{\infty }(D):={\bigcap }_{n=0}^{\infty }{C}^n(D)$ and is called the set of smooth functions.

Sum Rule

$$(f+g{)}^{(n)}(x_0)=f^{(n)}(x_0)+g^{(n)}(x_0)$$

Product Rule

$$(f\cdot g{)}^{(n)}(x_0)=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}(x_0)g^{(n-k)}(x_0)$$

Chain Rule

$$(g\circ f{)}^{\prime }(x_0)=g^{\prime }(f(x_0))f^{\prime }(x_0)$$

Quotient Rule

$${\left(\frac{f}{g}\right)}^{\prime }(x_0)=\frac{f^{\prime }(x_0)g(x_0)-f(x_0)g^{\prime }(x_0)}{g(x_0{)}^2}$$

Derivative of the Inverse

$${\left(f^{-1}\right)}^{\prime }(y_0)=\frac{1}{f^{\prime }(x_0)}$$

Rolle Theorem

Theorem

$a<b$, $f:[a,b]\to \mathbb{R}=$ a continuous and on $(a,b)$ differentiable function. If $f(a)=f(b)$: $⟹$ exists $\xi \in (a,b)$ with $f’(\xi )=0$

Intuition: We either have a constant line or one that bulges down or up, creating a minima or maxima.

Middle Value Theorem

Theorem

$a<b$, $f:[a,b]\to \mathbb{R}=$ a continuous and on $(a,b)$ differentiable function. $⟹$ exists $\xi \in (a,b)$ with $f’(\xi )=\frac{f(b)-f(a)}{b-a}$

Intuition: If a function is continuous and differentiable, then it’s derivative must change smoothly, taking on all values in its range.

Bernoulli-l’Hôpital

Theorem

Let $f,g$ be differentiable on an interval $I$, and let $x_0$ be a boundary point of $I$ (finite or $\pm \infty$). If i) $g(x)\ne 0$ and $g^{\prime }(x)\ne 0$ for all $x\in I$, ii) $\underset{x\to x_0}{\lim }f(x)=\underset{x\to x_0}{\lim }g(x)=0$ or $\underset{x\to x_0}{\lim }|f(x)|=\underset{x\to x_0}{\lim }|g(x)|=\infty$, iii) $L=\underset{x\to x_0}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$ exists (incl. $\pm \infty$), then $\underset{x\to x_0}{\lim }\frac{f(x)}{g(x)}=L$ also exists.

In short: under suitable conditions,

$$\lim \frac{f(x)}{g(x)}=\lim \frac{f^{\prime }(x)}{g^{\prime }(x)}=L$$

Version 1 — "$\frac{0}{0}$" at a finite point

$f,g:(a,b)\to \mathbb{R}$, limit $x\to a^{+}$, with $f,g\to 0$.

Example: $\underset{x\to 0^{+}}{\lim }\frac{\sin x}{x}=\underset{x\to 0^{+}}{\lim }\frac{\cos x}{1}=1$

Version 2 — "$\frac{\infty }{\infty }$" at a finite point

Same setup as Version 1, but with $|f|,|g|\to \infty$. $f$ may go to $+\infty$ or $-\infty$ — only the absolute value is required to diverge.

Example: $\underset{x\to 0^{+}}{\lim }\frac{\ln x}{1/x}$

Version 3 — limit at infinity

$f,g:(R,\infty )\to \mathbb{R}$, limit $x\to \infty$. Either form ($\frac{0}{0}$ or $\frac{\infty }{\infty }$) is allowed.

Example: $\underset{x\to \infty }{\lim }\frac{\ln x}{x}=\underset{x\to \infty }{\lim }\frac{1/x}{1}=0$

Warnings

• One-directional implication: $\lim \frac{f^{\prime }}{g^{\prime }}$ existing $⟹\lim \frac{f}{g}$ exists and equals it. Converse fails: $\frac{x+\sin x}{x}\to 1$ as $x\to \infty$, but $\frac{1+\cos x}{1}$ has no limit.
• Check the indeterminate form first. Applying the rule to a non-indeterminate quotient gives wrong answers.
• By symmetry, Versions 1 and 2 hold analogously for $x\to b^{-}$.

Convexity