Quotient rule provides a method to find the derivative of a function which is expressed in the form of quotient of two functions, like h(x) = $\frac{f(x)}{g(x)}$. Just as the derivative of a product function is not found by multiplying the derivatives of its factors, we can intuitively say that the derivative of a quotient is not the quotient of the derivatives. Let us verify this with an example.

Let f(x) = $x^3$. Using the power rule for derivatives we know f'(x) = $3x^2$. But if we choose to express f(x) as $\frac{x^{4}}{x}$, the derivative of $x^4$= $4x^3$ and the derivative of x = 1. The quotient of derivatives = $\frac{4x^3}{1}$ = $4^x^3$which is not the derivative of f(x). The quotient rule for derivatives is derived from the limit definition of differential coefficient by applying laws of limits.

Other than the quotient rule for derivatives, different quotient rules exist for different concepts like the quotient rule for exponents and logarithms, quotient rule for limits. Let us state and prove the quotient rule for derivatives and also solve few example problems.

## Proof

Quotient rule can be proved using the limit definition of derivative and limit laws.

$\frac{d}{dx}(\frac{u}{v})$ =  $\lim_{h->0}$  $\frac{\frac{u(x+h)}{v(x+h)} - \frac{u(x)}{v(x)}}{h}$ Limit definition of derivative

= $\lim_{h->0}$ $\frac{v(x)u(x+h)-v(x+h)u(x)}{h.v(x+h).v(x)}$ Complex fraction simplified.

Now to bring in the difference quotients corresponding the derivatives present in the formula, add and subtract v(x).u(x) in the denominator.

$\frac{d}{dx}(\frac{u}{v})$ = $\lim_{h \rightarrow 0}$ $\frac{v(x)u(x+h)-v(x)u(x)+v(x)u(x)-u(x)v(x+h)}{h.v(x+h)v(x)}$

= $\lim_{h->0}$ $\frac{v(x)(\frac{u(x+h)-u(x)}{h})-u(x)(\frac{v(x+h)-v(x)}{h})}{v(x+h)v(x)}$ Rearranged showing the difference quotients for the derivatives of u(x) and v(x).

= $\frac{\lim_{h->0}v(x)\lim_{h->0}\frac{u(x+h)-u}{h}-\lim_{h->0}u(x)\lim_{h->0}\frac{v(x+h)-v(x)}{h}}{\lim_{h->0}v(x+h)v(x)}$ Limit laws applied.

= $\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$ Limits applied and the quotient rule is proved.

## Quotient Rule for Exponents

Quotient Rule for exponents is one of the fundamental algebraic rules used to simplify and evaluate algebraic expressions.

$\frac{a^{m}}{a^{n}}$ = $a^{m-n}$ where m and n are real numbers. The exponent m - n is positive when m > n and negative when m < n.
Examples:
$5^7$ $\div$ $5^2$ = $\frac{5^{7}}{5^{2}}$ = $5^{7-2}$ = $5^5$

$x^3 \div x^9$ = $\frac{x^{3}}{x^{9}}$ = $x^{3-9}$= $x^{-6}$ = $\frac{1}{x^{6}}$

Some times the quotient rule for exponents is also stated as

$\frac{a^{m}}{a^{n}}$ = $a^{m-n}$ when m ≥ n and

$\frac{a^{m}}{a^{n}}$ = $\frac{1}{a^{n-m}}$ when m < n.

## Quotient Rule for Derivatives

Let u and v be differentiable functions and v(x) ≠ 0. Then the quotient $\frac{u}{v}$ is differentiable for x and $\frac{d}{dx}(\frac{u}{v})$ = $\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$.

The rule can be remembered using differentials as $\frac{v\ du - u\ dv}{v^{2}}$ and can be used for functions of any variable.
The rule can also be written using function notation as follows:
$\frac{d}{dx}[\frac{f(x))}{g(x)}]$ = $\frac{g(x)f'(x)-f(x)g'(x)}{g^{2}(x)}$

## Quotient Rule in Integration

A rule or method in integration is not generally used which corresponds to quotient rule in differentiation. But some experts modify the Integration by parts formula to produce a Quotient rule for integration.

Remember the integration by parts formula, which is considered as the counter part of product rule for derivatives
$\int U\ dV$ = uv - $\int V\ dU$

If the function u on the left side is a function in the denominator that is when we have a pattern like $\int vdu$ the integration by parts formula can be rewritten as a quotient rule for integration as follows:

$\int$ $\frac{dv}{u}$ =$\frac{v}{u}$ + $\int$ $\frac{v}{u^{2}}$ $du$
This formula can be derived from the parts formula by letting U = $\frac{1}{u}$ and V = v.

## Quotient Rule Examples

### Solved Examples

Question 1: Find the derivative of tan x using quotient rule.
Solution:

Let y = tan x
= $\frac{sinx}{cosx}$                     Quotient identity

Hence u(x) = sin x and v(x) = cos x.

u'(x) = cos x   and v'(x) = - sin x

$\frac{d}{dx}(\frac{u}{v})$ = $\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$                 Quotient Rule

$\frac{dy}{dx}$ = $\frac{cosx.(cosx)-sinx(-sinx))}{cos^{2}x}$                                     Substitution

= $\frac{cos^{2}x+sin^{2}x}{cos^{2}x}$                                               Simplified

= $\frac{1}{cos^{2}x}$                                                                        Pythagorean identity
= sec2 x                                                                                              Reciprocal identity
Thus $\frac{d}{dx}$tanx = sec2 x.

Question 2: Compute the derivative of f(x) = $\frac{x^{2}-5}{x+2}$

Solution:

u(x) = x2 - 5     and v(x) = x + 2
u'(x) = 2x and v'(x) = 1

f'(x) = $\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$                 Quotient Rule

= $\frac{(x+2)2x-(x^{2}-5)(1)}{(x+2)^{2}}$                        Substitution

= $\frac{2x^{2}+4x-x^{2}+5}{(x+2)^{2}}$

= $\frac{x^{2}+4x+5}{(x+2)^{2}}$

Question 3: Find the derivative of $\frac{2x^{3}-x^{2}+1}{x^{2}}$
Solution:

Here it is easier to find the derivative by splitting the terms of the numerator rather than applying quotient rule
y = $\frac{2x^{3}-x^{2}+1}{x^{2}}$ = 2x - 1 + x-2.

$\frac{dy}{dx}$ = 2 - 0 + (-2) x-3.                                               Applied the power formula for the derivatives.

= 2 - $\frac{2}{x^{3}}$