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
f(x) = m x + b
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
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
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
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
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}
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