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
\[ f(x) = \sqrt x \]
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
\(
f '(x) = \lim_{h\to\ 0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h\to\ 0} \dfrac{-1}{x(x+h)} = -\dfrac{1}{x^2}
\)
