# Use Definition to Find Derivative

## Definition of the First Derivative

Use the definition of the derivative to differentiate functions. This tutorial is well understood if used with the difference quotient .
The derivative f ' of function f is defined as
f'(x) = \lim_{h\to\ 0} \dfrac{f(x+h)-f(x)}{h}

when this limit exists. Hence, to find the derivative from its definition, we need to find the limit of the
difference quotient as h approaches zero.

## Use Definition to Find First Derivative - Examples with Solutions

Example 1
Use the definition of the derivative to find the derivative of function f defined by

f(x) = m x + b
where m and b are constants.
Solution to Example 1
We first need to calculate the difference quotient.
\dfrac{f(x+h)-f(x)}{h} = \dfrac{m(x+h)+b -(mx+b)}{h} \\\\ = \dfrac{mh}{h} = m
The derivative f ' is given by the limit of m (which is a constant) as h -->0. Hence
f'(x) = m
The derivative of a linear function f(x) = m x + b is equal to the slope m of its graph.

Example 2
Use the definition to find the derivative of

f(x) = a x^2 + bx + c

Solution to Example 2
We first find difference quotient
\dfrac{f(x+h)-f(x)}{h} = \dfrac{a(x + h)^2 + b(x + h) + c - ( a x^2 + b x + c )}{h}
Expand the expressions in the numerator and group like terms.
= \dfrac{a x^2 + 2 a x h + a h^2 + b x + b h + c - a x^2 - b x - c}{h}

= \dfrac{2 a x h + b h + a h^2}{h} = 2 a x + b + a h
The limit of 2 a x + b + a h as h -->0 is equal to 2 a x + b. Hence
f '(x) = 2 a x + b

Example 3
Find the derivative, using the definition, of function f given by

f(x) = sin x

Solution to Example 3
We first calculate the difference quotient
\dfrac{f(x+h)-f(x)}{h} = \dfrac{\sin (x + h) - \sin x }{h}
Use the trigonometric formula to transform a difference sin (x + h) - sin x in the numerator into a product.
\dfrac{\sin (x + h) - \sin x }{h} = \dfrac{2 cos [ (2 x + h)/2 ] sin (h/2)}{h}
Rewrite the above difference quotient as follows.
\dfrac{f(x+h)-f(x)}{h} = \dfrac{cos [ (2 x + h)/2 ] sin (h/2)}{h/2}
As h -->0, sin (h/2) / [ h / 2 ] --> 1 and cos [ (2 x + h)/2 ] --> cos (2x /2) = cos x. Hence the derivative of sin x is cos x
f '(x) = cos x

Example 4
Use the definition to differentiate

f(x) = √ x

Solution to Example 4
The difference quotient is given by
\dfrac{f(x+h)-f(x)}{h} = \dfrac{\sqrt{x+h} - \sqrt x}{h}

Multiply numerator and denominator by √ (x + h) + √ x , expand, group like terms and simplify.
= \dfrac{\sqrt{x+h} - \sqrt x}{h} \times \dfrac{\sqrt{x + h} + \sqrt x}{\sqrt{x + h} + \sqrt x}

= \dfrac{(\sqrt{x+h})^2- (\sqrt x)^2}{h(\sqrt{x + h} + \sqrt x)} = \dfrac{h}{h(\sqrt{x + h} + \sqrt x)} = \dfrac{1}{\sqrt{x + h} + \sqrt x}
As h -->0, 1 / [ √ (x + h) + √ x ] --> 1 / [ 2 √ (x) ] . Hence the derivative of √ x is 1 / [2 √ x]
f '(x) = \dfrac{1}{2\sqrt x}

Example 5
Use the definition to differentiate

f(x) = 1 / x

Solution to Example 5
The difference quotient is given by
\dfrac{f(x+h)-f(x)}{h} = \dfrac{1/(x+h) - 1/x}{h}

Set the two rational expressions in the numerator to the same denominator and rewrite the above as.
= \dfrac{\dfrac{x}{x(x+h)} - \dfrac{x+h}{x(x+h)}}{h}

= \dfrac{x-(x+h)}{x(x+h)h}

Simplify
= \dfrac{-1}{x(x+h)}

As h -->0, - 1 / [x(x+h)] --> - 1 / x2 . Hence the derivative of 1/ x is - 1 / x2
f '(x) = -\dfrac{1}{x^2}