# Difference between revisions of "Math 22 The Derivative and the Slope of a Graph"

## Slope of a Graph

We can estimate the slope at the given point to be

Slope = ${\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {\text{change in y}}{\text{change in x}}}}$

## Difference Quotient

 The slope ${\displaystyle m}$ of the graph of ${\displaystyle f}$ at the point ${\displaystyle (x,f(x))}$ can be
written as :

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

The right side of this equation ${\displaystyle {\frac {f(x+h)-f(x)}{h}}}$ is called Difference Quotient


Example: Find the Different Quotient of

1) ${\displaystyle f(x)=x^{2}-1}$

Solution: Consider ${\displaystyle {\frac {f(x+h)-f(x)}{h}}={\frac {(x+h)^{2}-1-(x^{2}-1)}{h}}}$

${\displaystyle ={\frac {x^{2}+2xh+h^{2}-1-x^{2}+1}{h}}={\frac {2xh+h^{2}}{h}}={\frac {h(2x+h)}{h}}=2xh}$

2) ${\displaystyle f(x)=4x-1}$

Solution:
Consider ${\displaystyle {\frac {f(x+h)-f(x)}{h}}={\frac {4(x+h)-1-(4x-1)}{h}}={\frac {4x+4h-1+4x+1}{h}}={\frac {4h}{h}}=4}$

## Definition of the Derivattive

 The derivative of ${\displaystyle f}$ at ${\displaystyle x}$ is given by

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

provided this limit exists. A function is differentiable at ${\displaystyle x}$ when its
derivative exists at . The process of finding derivatives is called differentiation.


Example: Use limit definition to find the derivative of

1) ${\displaystyle f(x)=x^{2}+2x}$

Solution: Consider: ${\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {(x+h)^{2}+2(x+h)-(x^{2}+2x)}{h}}}$

${\displaystyle =\lim _{h\to 0}{\frac {x^{2}+2xh+h^{2}+2x+2h-x^{2}-2x}{h}}=\lim _{h\to 0}{\frac {2xh+h^{2}+2h}{h}}}$

${\displaystyle \lim _{h\to 0}{\frac {h(2x+h+2)}{h}}=\lim _{h\to 0}(2x+h+2)=2x+(0)+2=2x+2}$

2) ${\displaystyle f(x)=2x^{2}-3x+1}$

Solution:
Consider: ${\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {2(x+h)^{2}-3(x+h)+1-(x^{2}-3x+1)}{h}}}$
${\displaystyle =\lim _{h\to 0}{\frac {2x^{2}+4xh+2h^{2}-3x-3h+1-2x^{2}+3x-1}{h}}=\lim _{h\to 0}{\frac {4xh+2h^{2}-3h}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {h(4x+2h-3)}{h}}=\lim _{h\to 0}(4x+2h-3)=4x+2(0)-3=4x-2}$