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
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}
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