Derivatives

Let $f(x)$ be a function defined in an open interval containing $a$. The derivative of a function $f(x)$ at $a$ is denoted by $f’(x)$ and is defined by:

$$f'(a) = \lim_{h\to0} {f(a+h) - f(a)\over h}$$

Differentiation Rules:

• Constant rule
• Power rule
• Sum rule
• Difference rule
• Constant multiple rule
• Product rule
• Quotient rule

Chain Rule If we have $h(x)=f(g(x))$ then $h’(x)=f’(g(x))g’(x)$

Derivatives of Exponentials & Logarithms

$${d \over dx}(b^{g(x)}) = b^{g(x)}g'(x)ln(b)$$

$${d \over dx}(log_b{g(x)}) = {g'(x) \over g(x)ln(b)}$$

Numerical Differentiation

• One-sided (forward) difference:

$${f(a+h) - f(a)\over h}$$

• One-sided (backward) difference:

$${f(a) - f(a-h)\over h}$$

• The central difference:

$${f(a+h) - f(a - h)\over 2h}$$

Fermat’s Theorem

If $f$ has a local extremum at $c$ and $f$ is differentiable at $c$, then $f’(c)=0$.

Extreme Value Theorem: A continuous function over a closed, bounded interval has an absolute maximum and minimum. Each extremum occurs at a critical point or an endpoint.

Rolle’s Theorem

Let $f$ be a continuous function over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$ such that $f(a) = f(b)$. There then exists at least one $c$ $\epsilon(a,b)$ such that $f’(c) = 0$.

Mean Value Theorem

Let $f$ be continuous over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$. Then, there exists at least one point $c$ $\epsilon(a,b)$, such that:

$$f'(c)={f(b)-f(a)\over b-a}$$

1. If $f’(x)>0$ for all $x$$\epsilon (a,b)$, then $f$ is an increasing function over $[a,b]$
2. If $f’(x)<0$ for all $x$$\epsilon (a,b)$, then $f$ is an decreasing function over $[a,b]$

Convex vs Concave Functions

Feature	Convex	Concave
Visual Shape	Cup / Smile ($\cup$)	Cap / Frown ($\cap$)
Midpoint Test	Line is above the curve	Line is below the curve
Optimization	Easy to find the minimum	Easy to find the maximum
Second Derivative	$f’‘(x) \ge 0$	$f’‘(x) \le 0$