Here's some basic information about derivatives:
 The derivative of a function
$\,f\,$ is a (new) function that gives the slopes of the
tangent lines to the graph of $\,f\,$.
Derivatives give slope information.
 The derivative of
$\,f\,$ is named $\,f'\,$ (using prime notation).
The function $\,f'\,$ is read aloud as ‘$\,f\,$ prime’.
 The
$\displaystyle\,\frac{d}{dx}\,$ operator (read as ‘dee dee x’) is an instruction to
‘take the derivative with respect to $\,x\,$ of whatever comes next’.
For example,
$\displaystyle\,\frac{d}{dx}(x^2)\,$ denotes the derivative with respect to $\,x\,$ of $\,x^2\,$.
Similarly, $\displaystyle\,\frac{d}{dt}(t^2)\,$ denotes the derivative with respect to $\,t\,$ of $\,t^2\,$.
The $\displaystyle\,\frac d{dx}\,$ operator is particularly useful when differentiating a function that has not been given a name.
If there is no confusion about the function that $\,\frac{d}{dx}\,$ is acting on, then the parentheses that hold the function
to be differentiated may be dropped.
 If a function $\,f\,$ is differentiable at $\,x\,$, then there is a nonvertical
tangent line at the point $\,\big(x,f(x)\bigr)\,$.
If there is a nonvertical tangent line at the point $\,\big(x,f(x)\bigr)\,$,
then the function $\,f\,$ is differentiable at $\,x\,$.
Thus, ‘$\,f\,$ is differentiable at $\,x\,$’ is equivalent to
‘there is a nonvertical tangent line to the graph of $\,f\,$ at the point $\,\bigl(x,f(x)\bigr)\,$’.
 The function $f'$, evaluated at $\,x\,$, is denoted by $\,f'(x)\,$.
‘$\,f'(x)\,$’ is read aloud as ‘$\,f\,$ prime of $\,x\,$’.
The number $\,f'(x)\,$ gives the slope of the tangent line to the graph of $\,f\,$ at the point $\,\bigl(x,f(x)\bigr)\,$.

Elaborating: for $\,f\,$ to be differentiable at $\,x\,$, all the following requirements must be met:
 $x$ must be in the domain of $f$, so the point $(x,f(x))$ exists
 there must be a tangent line to the curve at $(x,f(x))$;
the slope of this tangent line captures the ‘direction you're moving’ as you walk along the curve, going from
left to right
 the tangent line must be nonvertical, since a vertical line has no slope

Every important calculus idea is defined in terms of a limit.
For example, the derivative is defined in terms of a limit.

By definition, when the limit exists,
$\,\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)  f(x)}{h}\,$.

The expression $\displaystyle\,\frac{f(x+h)  f(x)}{h}\,$ gives the slope of the line between $\,\bigl(x,f(x)\bigr)\,$ and a nearby
point $\,\bigl(x+h,f(x+h)\bigr)\,$.
By taking the limit as $\,h\,$ approaches $\,0\,$, the nearby point is ‘slid’ closer and closer to the original point.

The definition of derivative is tedious to use, particularly as the function $\,f\,$ increases in complexity.
There must be a better way to find $\,f'(x)\,$!
The purpose of this lesson is to begin to develop shortcuts for finding derivatives.
 THE DERIVATIVE OF A CONSTANT IS ZERO
A constant function graphs as a horizontal line; the slope of a horizontal line is zero.
Examples:
 if $\,f(x) = 5\,$, then $\,f'(x) = 0\,$
 if $\,g(t) = \ln 3\,$, then $\,g'(t) = 0$
 $\displaystyle\frac{d}{dx}(\pi) = 0\,$, $\displaystyle\frac{d}{dt}(\sqrt{7}) = 0\,$, $\displaystyle\frac{d}{dw}({\text{e}}^3) = 0\,$
 THE DERIVATIVE OF A LINEAR FUNCTION IS THE SLOPE OF THE LINE
Recall that the graph of $\,f(x) = mx + b\,$ is a line with slope $\,m\,$.
Thus, $\,f'(x) = m\,$.
Examples:

if $g(t) = 5t  1\,$, then $\,g'(t) = 5\,$

$\displaystyle\frac{d}{dx} (3  \frac{2}{7}x) = \frac 27\,$;
note that $\,y = 3  \frac 27x = \frac 27x + 3\,$ graphs as a line with slope $\,\frac 27\,$.
SIMPLE POWER RULE
how to differentiate $\,x^n$
For all real numbers $\,n\,$:
$$\frac d{dx} x^n = nx^{n1}$$
NOTES ON THE SIMPLE POWER RULE:
 Power functions are functions that can be written in the form $\,x^n\,$, for some real number $\,n\,$.
Examples of power functions:
 $x^2\,$, $x^3\,$, $x^{1.4}\,$, $x^\pi$
 $\displaystyle\frac 1x = x^{1}\,$, $\displaystyle\frac 1{x^2} = x^{2}\,$, $\sqrt x = x^{1/2}\,$,
$\root 3\of {x^2} = x^{2/3}$
The Simple Power Rule tells how to differentiate power functions.
 To differentiate $\,x^n\,$:
 bring the exponent down front;
 decrease the original exponent by $\,1$
 EXAMPLES:
 if $f(x) = x^3\,$, then $f'(x) = 3x^2$
 $\displaystyle \frac{d}{dx} \sqrt x = \frac d{dx} x^{1/2} =
\frac{1}{2} x^{1/2} = \frac{1}{2x^{1/2}} = \frac{1}{2\sqrt x}$
 $\displaystyle \frac{d}{dt} \frac{1}{\sqrt t} = \frac d{dt} t^{1/2} =
\frac{1}{2} t^{3/2} = \frac{1}{2t^{3/2}} = \frac{1}{2\sqrt {t^3}}$
 $\displaystyle \frac{d}{dt} t\root 3\of t = \frac{d}{dt} t^1 t^{1/3} = \frac d{dt} t^{4/3}
= \frac 43 t^{1/3} = \frac 43\root 3\of t = \frac {4\root 3\of t}{3}$
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Differentiation Formula Practice
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.