Calculus/Differentiation

Prerequisites

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

Calculus/Limits

Geometry

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
Addition Rule	f(x)+g(x)	f'(x)+g'(x)
???	x	1
Scalar Rule	a•f(x)	a•f'(x)
Power Rule	(f(x))a	a•(f(x))a-1ˆ•f'(x)
Product Rule	f(x)•g(x)	f(x)•g'(x) + g(x)•f'(x)
Quotient Rule	f(x)/g(x)	(g(x)•f'(x) - f(x)•g'(x)) / (g(x))2
derivative of trigonometric function	sin(x)	cos(x)
derivative of trigonometric function	cos(x)	-sin(x)
derivative of trigonometric function	tan(x)	sec2(x)
derivative of trigonometric function	ln(x)	1/x
derivative of trigonometric function	ex	ex