The derivative f ' of function f is defined as
f '(x) = lim_{ h >0} [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
Solution to Example 1
 We first need to calculate the difference quotient.
[ f (x + h) – f(x) ] / h = [ m (x + h) + b – (m x + b) ] / h
= [ m h ] / h = m
 The derivetive f ' is given by the limit of m 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) = ax^{ 2} + bx + c
Solution to Example 2
 We first find difference quotient
[ f (x + h) – f(x) ] / h =
[ a(x + h)^{ 2} + b(x + h) + c  ( ax^{ 2} + bx + c ) ] / h
 Expand the expressions in the numerator and group like terms.
= [ ax^{ 2} + 2 a x h + a h^{ 2} + bx + bh + c  ax^{ 2}  bx  c] / h
= [ 2 a x h + bh + 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
[ f (x + h) – f(x) ] / h =
[ sin (x + h)  sin x ] / h
 Use the trigonometric formula to transform a difference sin (x + h)  sin x into a product.
[ sin (x + h)  sin x ] / h = 2 cos [ (2 x + h)/2 ] sin (h/2) / h
 Rewrite the above difference quotient as follows.
[ f (x + h) – f(x) ] / h = 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 deinition to differentiate
f(x) = sqrt x
Solution to Example 4
 The difference quotient is given by
[ f (x + h) – f(x) ] / h =
[ sqrt (x + h)  sqrt x ] / h
 Mutliply numerator and denominator by [ sqrt (x + h) + sqrt x ], expand, group like terms and simplify.
= [ sqrt (x + h)  sqrt x ][ sqrt (x + h) + sqrt x ] / [ h ( sqrt (x + h) + sqrt x ) ]
= h / [ h ( sqrt (x + h) + sqrt x ) ]
= 1 / [ sqrt (x + h) + sqrt x ]
 As h >0, 1 / [ sqrt (x + h) + sqrt x ] > 1 / [ 2 sqrt (x) ] . Hence the derivative of sqrt x is 1 / [2 sqrt x]
f '(x) = 1 / [2 sqrt x]
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