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}

Use Definition to Find Derivative

Definition of the First Derivative

Use Definition to Find First Derivative - Examples with Solutions

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