Differentiability

A function $f$ is differentiable at $a$ iff:

$$\lim_{x\to a}{\frac{f(x)-f(a)}{x-a}} = L \in \mathbb{R} = f'(a)$$

When it is differentiable, $f'(a)$ is called the derivative of $f$ at $a$.

Critical point

$c\in[a,b]$ is called a critical point iff: $$

$$f\text{ is not differentiable at } c \;\; \lor \;\; f'(c)=0$$

One-side differentiable

$f$ is differentiable at $a$ iff $f$ is left differentiable at $a$ and $f$ is right differentiable at $a$.

Left differentiable

A function $f$ is left-differentiable at $a$ iff:

$$\lim_{x\to a^{-}}{\frac{f(x)-f(a)}{x-a}} = L \in \mathbb{R} = f'_{-}(a)$$

Right differentiable

A function $f$ is right-differentiable at $a$ iff:

$$\lim_{x\to a^{+}}{\frac{f(x)-f(a)}{x-a}} = L \in \mathbb{R} = f'_{+}(a)$$

On intervals

Open interval

A function $f$ is differentiable in $(a,b)$ iff $f$ is differentiable on every $c\in(a,b)$.

Closed interval

A function $f$ is differentiable in $[a,b]$ iff $f$ is:

• differentiable on every $c\in(a,b)$
• right-differentiable at $a$
• left-differentiable at $b$

Continuously differentiable functions

A function $f$ is said to be continously differentiable at $a$ iff :

• $f$ is differentiable at $a$ and
• $f'$ is continous at $a$

Differentiability implies continuity

$f \text{ is differentiable at } a \implies f \text{ is continuous at } a$

Likewise, one-sided differentiability implies corresponding one-sided continuity.

Proof Hint

Use $\delta=min(\delta_1, \frac{\epsilon}{1 + |f'(a)|})$. $$

Note

Suppose $f$ is differentiable at $a$. Define $g$:

$$g(x) = \begin{cases} \; \cfrac{f(x)-f(a)}{x-a}, & x \neq a \\[2ex] \; f'(a), & x=a \end{cases}$$

$g$ is continuous at $a$.

Properties of differentiation

Addition

$$\frac{\text{d}}{\text{d}x} (f \pm g) = f' \pm g'$$

Multiplication

$$\frac{\text{d}}{\text{d}x} (fg) = fg' + fg'$$

Division

$$\frac{\text{d}}{\text{d}x} \bigg(\frac{f}{g}\bigg) = \frac{gf' - fg'}{g^2}$$

Composition

$$\frac{\text{d}}{\text{d}x} f(g(x)) = f'(g(x))\,g'(x)$$

Power

$$\frac{\text{d}}{\text{d}x} f^n = nf^{n-1}(x) f'(x)$$