# Use Definition to Find Derivative

## First Derivative

Use the definition of the derivative to differentiate functions. This tutorial is well understood if used with the difference quotient and the definition of the derivative .
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.

### 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 h}{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}