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.


Examples with Detailed 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.
\( \dfrac{f(x+h)-f(x)}{h} = \dfrac{m(x+h)+b -(mx+b)}{h} \)
Simplify
\( = \dfrac{m h}{h} = m \)
The derivative \( f '\) is given by the limit of \( m \) (which is a constant) as \( {h\to\ 0} \). Hence
\( f'(x) = \lim_{h\to\ 0} \dfrac{f(x+h)-f(x)}{h}= \lim_{h\to\ 0} m = m \)
The derivative of a linear function \( f(x) = m x + b \) is equal to the slope \( m \) of its graph which is a line.



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.
\( = \dfrac{a x^2 + 2 a x h + a h^2 + b x + b h + c - a x^2 - b x - c}{h} \)
Simplify.
\( = \dfrac{2 a x h + b h + a h^2}{h} = 2 a x + b + a h \)
The derivative of \( f(x) = a x^2 + bx + c \) is given by the limit of the difference quotient. Hence
\( f '(x) = \lim_{h\to\ 0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h\to\ 0} (2 a x + b + a h) = 2 a x + b \)



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.
\( \dfrac{f(x+h)-f(x)}{h} = \dfrac{\cos [ (2 x + h)/2 ] \sin (h/2)}{h/2} \)
The derivative is given by the limit of the difference quotient. Hence
\( f '(x) = \lim_{h\to\ 0} \dfrac{f(x+h)-f(x)}{h} \\ = \lim_{h\to\ 0} \dfrac{\cos [ (2 x + h)/2 ] \sin (h/2)}{h/2} \)
Use the theorems of the limit of the product of two functions to write
\( f '(x) = \lim_{h\to\ 0} \dfrac{cos [ (2 x + h)/2 ] \sin (h/2)}{h/2} = \lim_{h\to\ 0} cos [ (2 x + h)/2 ] \times \lim_{h\to\ 0} \dfrac{\sin (h/2)}{h/2} \)
The limits in the above product are given by
\( \lim_{h\to\ 0} \cos [ (2 x + h)/2 ] = \cos (2 x / 2) = \cos x \)
and
\( \lim_{h\to\ 0} \dfrac{\sin (h/2)}{h/2} = \lim_{t\to\ 0} \dfrac{\sin (t)}{t} = 1 \)
The derivative of \( f(x) = \sin x \) is given by the limit of the difference quotient. Hence
\( f '(x) = \lim_{h\to\ 0} \dfrac{f(x+h)-f(x)}{h} = \cos x \times 1 = \cos x \)



Example 4
Use the definition to differentiate
\[ f(x) = \sqrt x \]


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 \( \sqrt{x + h} + \sqrt{x} \), expand, group like terms and simplify.
\( = \dfrac{\sqrt{x+h} - \sqrt x}{h} \times \dfrac{\sqrt{x + h} + \sqrt x}{\sqrt{x + h} + \sqrt x} \)
Expand and group.
\( = \dfrac{(\sqrt{x+h})^2- (\sqrt x)^2}{h(\sqrt{x + h} + \sqrt x)} = \dfrac{(x+h) - x}{h(\sqrt{x + h} + \sqrt x)} = \dfrac{h}{h (\sqrt{x + h} + \sqrt x)} \)
Cancel \( h \) and simplify.
\( = \dfrac{1}{\sqrt{x + h} + \sqrt x} \)
The derivative of \( f(x) = \sqrt x \) is given by the limit of the difference quotient. Hence
\( f '(x) = \lim_{h\to\ 0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h\to\ 0} \dfrac{1}{\sqrt{x + h} + \sqrt x} = \dfrac{1}{2\sqrt x} \)



Example 5
Use the definition to differentiate
\[ f(x) = \dfrac{1}{x} \]
Solution to Example 5
The difference quotient is given by
\( \dfrac{f(x+h)-f(x)}{h} = \dfrac{\dfrac{1}{x+h} - \dfrac{1}{x}}{h} \)
Set the two rational expressions in the numerator to the same denominator and rewrite the above as.
\( = \dfrac{\dfrac{x}{x(x+h)} - \dfrac{x+h}{x(x+h)}}{h} \)
which simplifies to.
\( = \dfrac{x-(x+h)}{x(x+h)h} \)
\( = \dfrac{-1}{x(x+h)} \)
The derivative of \( f(x) = \dfrac{1}{x} \) is given by the limit of the difference quotient. Hence
\( f '(x) = \lim_{h\to\ 0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h\to\ 0} \dfrac{-1}{x(x+h)} = -\dfrac{1}{x^2} \)


More Links and References

difference quotient
differentiation and derivatives
Difference Quotient Calculator

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