# Product Rule of Differentiation with Examples

 

The steps to prove the product rule of differentiation are presented along with examples, exercises and solutions.

## Definition of the Derivative of a Function

The derivative $f'(x)$ of the function $f(x)$ is defined as
$f'(x) = \lim_{h \to 0} \; \dfrac{f(x+h) - f(x)}{h} \qquad (1)$

## Derivative of the Product of two Functions

Let function $f(x)$ be given by the product of two functions $u(x)$ and $v(x)$ written as
$\quad f(x) = u(x) v(x)$
Using the definition of the derivative in $(1)$, the derivative of $f(x) = u(x) v(x)$ is given by

$\qquad \displaystyle f'(x) = \lim_{h \to 0} \; \dfrac{u(x+h) v(x+h) - u(x) v(x)}{h} \qquad (2)$

Subtracting and adding the same quantity to the denominator of $(2)$ above does not change it.
Subtract and add $u(x) v(x+h)$ in the numerator of $(2)$ above and write

$\qquad \displaystyle f'(x) = \lim_{h \to 0} \; \dfrac{u(x+h) v(x+h) \color{red}{- u(x) v(x+h)} - u(x) v(x) \color{red}{+ u(x) v(x+h)} }{h} \qquad (3)$

Split $(3)$ and rewrite it as follows
$\qquad \displaystyle f'(x) = \lim_{h \to 0} \left( \; \dfrac{u(x+h) v(x+h) - u(x) v(x+h)}{h} + \dfrac{u(x) v(x+h) - u(x) v(x) }{h} \right) \qquad (4)$

From the properties of limits , the limit of a sum is equal to the sum of the limits, hence
$\qquad \displaystyle f'(x) = \lim_{h \to 0} \; \dfrac{u(x+h) v(x+h) - u(x) v(x+h)}{h} + \lim_{h \to 0} \left ( \dfrac{u(x) v(x+h) - u(x) v(x) }{h} \right) \qquad (5)$

Use factoring to rewrite the above as
$\qquad \displaystyle f'(x) = \lim_{h \to 0} \; v(x+h) \dfrac{u(x+h) - u(x) }{h} + \lim_{h \to 0} u(x) \dfrac{ v(x+h) - v(x) }{h} \qquad (6)$

From the properties of limits , the limit of a product is equal to the product of the limits, hence the above may be written as
$\qquad \displaystyle f'(x) = \lim_{h \to 0} \; v(x+h) \lim_{h \to 0} \; \dfrac{u(x+h) - u(x) }{h} + \lim_{h \to 0} u(x) \lim_{h \to 0} \; \dfrac{ v(x+h) - v(x) }{h} \qquad (7)$

Evaluate the individual limits in $(7)$
$\qquad \displaystyle \lim_{h \to 0} \; v(x+h) = v(x)$
$\qquad \displaystyle \lim_{h \to 0} u(x) = u(x)$
$\qquad \displaystyle \lim_{h \to 0} \; \dfrac{u(x+h) - u(x) }{h} = u'(x)$ , according to the definition of the derivative given in $(1)$.
$\qquad \displaystyle \lim_{h \to 0} \; \dfrac{ v(x+h) - v(x) }{h} = v'(x)$ , according to the definition of the derivative given in $(1)$.

Substitute the individual limits above in $(7)$ to obtain
$\qquad \displaystyle f'(x) = u'(x) v(x) + u(x) v'(x)$

Hence the rule of the derivative of a product is given by $(u v)' = u'(x) v(x) + u(x) v'(x) \qquad (I)$

## Examples with Solutions

### Example 1

Find the derivatives of
a) $\quad f(x) = x \; \ln(x)$       b) $\quad g(x) = \sin(x) e^x$

Solution
a)
Let $u(x) = x$ and $v(x) = \ln x$ and write $f(x)$ as the product of $u$ and $v$ as follows
$\quad f(x) = u(x) \; v(x)$

Write that $\quad f'(x) = (u(x) \; v(x))'$

Use the product rule in $(I)$
$\quad f'(x) = (u(x) \; v(x))' = u'(x) v(x) + u(x) v'(x) \quad (8)$

Calculte the derivatives $u'$ and $v'$
$\quad u'(x) = 1$ and $v' = \dfrac{1}{x}$

Substitute $u, u', v, v'$ by their expressions in $(8)$ above to obtain $\quad f'(x) = 1 \cdot \ln x + x \cdot \dfrac{1}{x}$

Simplify to obtain the final answer as $\quad f'(x) = \ln x + 1$

b)
Let $w(x) = \sin(x)$ and $z(x) = e^x$ and write $g(x)$ as the product of $w$ and $z$ as follows
$\quad g(x) = w(x) \; z(x)$

Write that
$\quad g'(x) = ( w(x) \; z(x) )'$

Use the product rule in $(I)$
$\quad g'(x) = (w(x) \; z(x))' = w'(x) z(x) + w(x) z'(x) \quad (9)$

Calculte the derivatives $w'$ and $z'$
$\quad w'(x) = cos(x)$ and $z'(x) = e^x$

Substitute $w, w', z, z'$ by their expressions in $(9)$ above to obtain
$\quad g'(x) = cos(x) \cdot e^x + sin(x) \cdot e^x$

Factor $e^x$ to obtain the final answer as
$\quad g'(x) = \left( cos(x) + sin(x) \right) \; e^x$

### Example 2

Calculate the derivative of $h(x) = (2x+3) \cos (x) \ln x$

Solution
Function $h(x)$ is given by the product of three functions. Let $u = 2x+3$, $v = \cos(x)$ and $w = \ln x$ and write $h(x)$ as
$\quad h(x) = u(x) v(x) w(x)$

Use the product rule in $I$ to write
$\quad h'(x) = u(x) (v(x) w(x))' + u'(x) (v(x) w(x))$

Use the product rule in $(I) m$ one more time to the term $(v(x) w(x))'$ and write
$\quad h'(x) = u(x) (v'(x) w(x) + v(x) w'(x) ) + u'(x) (v(x) w(x))$

Expand and write a form that is easy to retain

$\quad h'(x) = \color{red}{u'(x)} v(x) w(x) + u(x) \color{red}{v'(x)} w(x) + u(x) v(x) \color{red}{w'(x)} \quad (10)$

Evaluate the derivatives of $u, v , w$ $u'(x) = 2$ , $v'(x) = - \sin(x)$ , $w' = \dfrac{1}{x}$

Substitute $u, u', v, v', w, w')$ by their expressions in $(3)$ above to obtain
$\quad h'(x) = 2 \cos(x) \ln x - (2x+3) \sin(x) \ln x + \dfrac{(2x+3) \cos(x)}{x}$

## Exercises

Find the derivatives of the functions

1. $\quad f(x) = (3x-5) \cos(x)$
2. $\quad g(x) = (-4x+3) e^x$
3. $\quad h(x) = x^3 \sin(x) e^x$

## Solutions to the Above Exercises

1. $\quad f'(x) = 3\cos (x) - (3x-5) \sin (x)$
2. $\quad g'(x) = -4e^x + (-4x+3)e^x = (-4 x - 1) e^x$
3. $\quad h'(x) = 3x^2\sin (x)e^x + x^3 \cos (x)e^x + x^3 \sin(x) e^x = 3x^2\sin (x)e^x + (\cos (x) + \sin(x) ) x^3 e^x$