# Quotient Rule of Differentiation with Examples

 

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

## Derivative of the Quotient of two Functions

Let function $f(x)$ be given by the quotient of two functions $u(x)$ and $v(x)$ written as
$\quad f(x) = \dfrac{u(x)}{ v(x)} \qquad (1)$

Mutliply both sides of the above by $v(x)$ and simplify to obtain
$\quad f(x) v(x) = u(x)$

Take the derivative of both sides of the above
$\quad ( f(x) v(x))' = u'(x)$

Apply product rule of differentiation to the left side
$\quad f'(x) v(x) + f(x) v'(x) = u'(x)$

Solve the above for $f'(x)$
$\quad f'(x) = \dfrac{ u'(x) - f(x) v'(x) }{ v(x) }$

Substitute $f(x)$ by $\dfrac{u(x)}{ v(x)}$ as given in $(1)$ above

$\quad f'(x) = \dfrac{ u'(x) - \dfrac{u(x)}{ v(x)} v'(x) }{ v(x) }$

Rewrite the expression in the numerator with a common denominator and simplify to rewrite $f'(x)$ as

$\quad f'(x) = \dfrac{ u'(x) v(x) - u(x) v'(x) }{ (v(x))^2 }$

Finally the quotient rule of differentiation
$\left(\dfrac{u}{v}\right)' = \dfrac{ u'(x) v(x) - u(x) v'(x) }{ (v(x))^2 } \qquad (I)$

## Examples with Solutions

### Example 1

Find the derivatives of
a) $\quad f(x) = \dfrac{x}{\ln x}$       b) $\quad g(x) = \dfrac{\sin(x)}{\cos(x)}$

Solution
a)
Let $u(x) = x$ and $v(x) = \ln x$ and write $f(x)$ as the quotient of $u$ and $v$ as follows
$\quad f(x) = \dfrac {u} {v}$

Use the quotient rule in $(I)$ to write the derivative $f'(x)$ as
$\quad f'(x) = \dfrac{u'(x) v(x) - u(x) v'(x) }{(v(x))^2} \qquad (2)$

Given $u(x) = x$ and $v(x) = \ln x$, the derivatives $u'$ and $v'$ are given by
$\quad u' = 1$ and $v' = \dfrac{1}{x}$

Substitute $u, u', v, v'$ by their expressions in $(2)$ above to obtain
$\quad f'(x) = \dfrac{ 1 \cdot \ln x - x \cdot \dfrac{1}{x} }{ \ln^2 x }$

Simplify to obtain

$\quad f'(x) = \dfrac{ \ln x - 1} { \ln^2 x }$

b)
Let $w(x) = \sin x$ and $z(x) = \cos x$ and write $g(x)$ as the quotient of $w$ and $z$ as follows
$\quad f(x) = \dfrac {w} {z}$

Use the quotient rule in $(I)$ to write the derivative $g'(x)$ as
$\quad g'(x) = \dfrac{w'(x) z(x) - w(x) z'(x) }{(z(x))^2} \qquad (3)$

Given $w(x) = \sin x$ and $z(x) = \cos x$, the derivatives $w'$ and $z'$
$\quad w' = \cos x$ and $z' = - \sin x$

Substitute $w, w', z, z'$ by their expressions in $(3)$ above to obtain
$\quad g'(x) = \dfrac{ \cos x \cos x - \sin x (- \sin x) }{ \cos^2 x }$

The numerator of the above is equal to $\cos x \cos x - \sin x (- \sin x) = \cos^2 x + \sin^2 x = 1$, hence simplify to obtain
$\quad g'(x) = \dfrac{ 1 }{ \cos^2 x }$
Use the identity $\sec x = \dfrac{1}{\cos x}$ to write the final answer as
$g'(x) = \sec^2 x$

### Example 2

Calculate the derivative of $h(x) = \dfrac{x \; \ln x}{\sin x \; e^x}$

Solution
Let $u(x) = x \; \ln x$ and $v(x) = \sin x \; e^x$ and write $h(x)$ as the quotient
$h(x) = \dfrac{u(x)}{v(x)}$

Use the quotient rule in $(I)$ to write the derivative $h'(x)$ as
$\quad h'(x) = \dfrac{ u'(x) v(x) - u(x) v'(x) }{ (v(x))^2 } \qquad (4)$

Given $u(x) = x \; \ln x$ which is the product of two functions, the derivative $u'$ is calculated using the product rule of differentiation
$u'(x) = (x)'\ln x + x (\ln x)' = \ln x + x \dfrac{1}{x} = \ln x + 1$

$v(x) = \sin x \; e^x$ is also the product of two functions, the derivative $v'$ is calculated using the product rule of differentiation
$v'(x) = (\sin x)'e^x + \sin x (e^x)' = \cos x e^x + \sin x e^x = (\cos x + \sin x) e^x$

Substitute $u, u', v, v'$ by their expressions in $(4)$ above to obtain
$\quad h'(x) = \dfrac{ (\ln x + 1) (\sin x \; e^x) - (x \; \ln x )((\cos x + \sin x) e^x) }{ (\sin x \; e^x )^2 }$
which may be rewritten as
$\quad h'(x) = \dfrac{ (\ln x + 1) \sin x - x \; \ln x \; (\cos x + \sin x) }{ \sin^2 x \; e^{x} }$

## Exercises

Find the derivatives of the functions

1. $\quad f(x) = \dfrac{\cos(x)}{\sin x}$

2. $\quad g(x) = \dfrac{-2x+4}{e^x}$

3. $\quad h(x) = \dfrac{ 2x \; e^x}{ x \sin x}$

## Solutions to the Above Exercises

1. $\quad f'(x) = \dfrac{-\sin x \sin x - \cos x \cos x}{\sin^2 x} = \dfrac{-(\sin^2 x + \cos^ x)}{\sin^2 x} = \dfrac{-1}{\sin^2 x} = - \csc^2 x$

2. $\quad g'(x) = \dfrac{-2 \; e^x - (-2x+4) \; e^x}{ (e^x)^2} = -\dfrac{2\left(-x+3\right)}{e^x}$

3. $\quad h(x) = \dfrac{ (2(1+x)\; e^x) (x \sin x) - (2x \; e^x) ( \sin x + x \; \cos x)}{ (x \; \sin x)^2} \\ \qquad \qquad = \dfrac{ 2 x \sin x \; e^x + 2 x^2 \sin x \; e^x - 2 x \sin x \; e^x - 2 x^2 \cos x \; e^x} { (x \; \sin x)^2} \\ \qquad \qquad = 2\left(\csc \left(x\right)-\cot \left(x\right)\csc \left(x\right)\right) e^x$