# Calculus/Differentiation

## Prerequisites

In order to understand and calculate derivatives, one must understand the following topics:

## Introduction to Differentiation

Differentiation is the process of finding the unique slope of a line at any point on that line, should such a slope exist. As we shall show, there are commonplace functions that fail this requirement at some point.

Generally, a slope can be found by taking the change in y-coordinates divided by the change in x-coordinates or:

${\frac {(y_{2}-y_{1})}{(x_{2}-x_{1})}}={\frac {\Delta y}{\Delta x}}=m$

where m is the slope.

Now what happens when we attempt to take the slope of a point on a line, or how fast the y's are changing in respect to the x's at a single point? The formula fails when $\Delta x$  or $\Delta y=0$

In order to do this, we have to undergo differentiation.

## Definition of a Derivative

The following formula will calculate the slope of a line at any point:

$f^{\prime }(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

where $f^{\prime }$  indicates the derivative, and h is the difference between a point h units away from x and x

Given $f(x)=3x^{2}$ , to find the derivative of f(a) (where a is any x coordinate within the domain of f(x)), use the definition of derivative.

$f^{\prime }(x)=\lim _{h\to 0}{\frac {3(a+h)^{2}-3a^{2}}{h}}$

Here's an exercise to try:

Find $f^{\prime }(2)$ , given that $f(x)=4x^{2}$ . The solution is below.

$f^{\prime }(2)=\lim _{h\to 0}{\frac {4(2+h)^{2}-16}{h}}=\lim _{h\to 0}{\frac {4(4+4h+h^{2})-16}{h}}=\lim _{h\to 0}{\frac {16+16h+4h^{2}-16}{h}}=\lim _{h\to 0}{\frac {16h+4h^{2}}{h}}=\lim _{h\to 0}(16+4h)=16$

Therefore, the slope of 4x^2 at x=2 is 16.

This is a somewhat tedious process when bigger functions are involved. Take for example:

$f(x)=3x^{6}+sin^{2}x-12^{x}+5$

Using the definition of derivative, your equation looks like this:

$f^{\prime }(x)=\lim _{h\to 0}{\frac {(3(x+h)^{6}+sin^{2}(x+h)-12^{(x+h)}+5)-(3x^{6}+sin^{2}(x)-12^{x}+5)}{h}}$

Have fun solving that algebraically!

Luckily, this is where rules for derivatives come in.

## Where Derivation Fails

Two examples of commonplace functions either have ambiguous derivatives or none at some point. The absolute value function has an ambiguous derivative at x=0 For |x|, |x| = -x and the derivative of |x| is -1 if x < 0. |x| = x and the derivative of |x| is 1 if x > 0. But if x=0, then |x|, x, and -x are all zero. No matter how close a number is to zero, the derivative of |x| is 1 for x > 0 but -1 for any x less than zero. Because one cannot determine whether the derivative of |x| is -1, 1, or something else at x=0, one has an ambiguous derivative at x=0... and thus none.

Another function that most off us know well is 1/x. For f(x) = 1/x

$f^{\prime }(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$  implies

$f^{\prime }(1/x)=\lim _{h\to 0}{\frac {f(1/(x+h))-f(1/x)}{h}}$

$f^{\prime }(1/x)=\lim _{h\to 0}{\frac {(x/x(x+h))-(x+h)/x(x+h)}{h}}$

$f^{\prime }(1/x)=\lim _{h\to 0}{\frac {(x-(x+h))}{hx(x+h)}}$

$f^{\prime }(1/x)=\lim _{h\to 0}{\frac {-h}{hx(x+h)}}$

$f^{\prime }(1/x)=\lim _{h\to 0}{\frac {-1}{x(x+h)}}$

$f^{\prime }(1/x)=\lim _{h\to 0}{\frac {-1}{x^{2}}}$

$f^{\prime }(1/x)={\frac {-1}{x^{2}}}$

No derivative of 1/x can exist at x=0.

## Derivative Rules

In the following table, f(x) and g(x) are functions whereas "a" is a constant number. f'(x) and g'(c) are the derivatives of the functions f(x) and g(x), respectively.

Rule Name Original Function Derivative
Chain Rule f(g(x)) f'(g(x))•g'(x)
constant function derivative a 0