 # Quotient Rule

In Calculus, the Quotient Rule is a method for determining the derivative (differentiation) of a function which is the ratio of two functions that are differentiable in nature. It is a formal rule used in the differentiation problems in which one function is divided by the other function. The quotient rule follows the definition of the limit of the derivative. Always remember that the quotient rule always begins with the bottom function and it ends with the bottom function squared.  In this article, we are going to have a look at the definition, quotient rule formula, proof and examples in detail.

## Quotient Rule Definition

In Calculus, a Quotient rule is similar to the product rule. A Quotient Rule is stated as the ratio of the quantity of the denominator times the derivative of the numerator function minus the numerator times the derivative of the denominator function to the square of the denominator function. In short, quotient rule is a way of differentiating the division of functions or the quotients.

## Quotient Rule Formula

Let the given function be f(x), which is given by:

$\large \mathbf{f(x) = \frac{s(x)}{t(x)}}$,

Thus the differentiation of the function is given by:

$\large \mathbf{f'(x) = \left [ \frac{s(x)}{t(x)} \right ]’ = \frac{t(x).s'(x) – s(x). t'(x)}{\left \{ t(x) \right \}^{2}}}$

The quotient rule of differentiation is defined as the ratio of two function (1st function / 2nd Function), is equal to the ratio of (Differentiation of 1st function $\large \times$ the 2nd function – Differentiation of second function $\large \times$ the 1st function) to the square of the 2nd function.

## Quotient Rule Proof

We know, the derivative of a function is given as:

$\large \mathbf{f'(x) = \lim \limits_{h \to 0} \frac{f(x+h)- f(x)}{h}}$

Thus the derivative of ratio of function is: Hence, the quotient rule is proved.

### Quotient Rule Example

Let us work out some examples:

Example 1:

Find the derivative of $\tan x$.

Solution:

We know, $\tan x = \frac{\sin x}{\cos x}$

$\left (\tan x \right )’ = \frac{\mathrm{d} }{\mathrm{d} x} \left (\frac{\sin x}{\cos x} \right )$

$= \left ( \frac{\cos x . (\sin x)’ – \sin x (\cos x)’}{\cos^{2}x} \right )$

$= \left ( \frac{\cos^{2} x + \sin^{2} x }{\cos^{2}x} \right )$

$= \left ( \frac{1}{\cos^{2}x} \right )$$= \sec^{2} x$

Example 2:

Find the derivative of $\sqrt{\frac{5x + 7}{3x – 2}}$

Solution:

$\sqrt{\frac{5x + 7}{3x – 2}} = \frac{\sqrt{5x + 7}}{\sqrt{3x – 2}}$

Applying the quotient rule, we have

$\frac{\mathrm{d} }{\mathrm{d} x}\left (\sqrt{\frac{5x + 7}{3x – 2}} \right ) = \frac{\sqrt{3x – 2}. \frac{\mathrm{d} }{\mathrm{d} x}\left (\sqrt{5x + 7} \right ) – \sqrt{5x + 7} . \frac{\mathrm{d} }{\mathrm{d} x} \sqrt{3x – 2} }{3x – 2}$

$= \frac{\sqrt{3x – 2}. \left (\frac{5}{2.\sqrt{5x + 7}} \right ) – \sqrt{5x + 7} . \left (\frac{3}{2.\sqrt{3x – 2}} \right ) }{3x – 2}$

$= \frac{\left (\frac{5.\sqrt{3x – 2}}{2.\sqrt{5x + 7}} \right ) – \left (\frac{3. \sqrt{5x + 7}}{2.\sqrt{3x – 2}} \right ) }{3x – 2}$

Taking LCM, we have

$= \frac{5.\left (3x – 2 \right ) – 3. \left (5x + 7 \right )}{2\left (3x – 2 \right )\left ( \sqrt{3x – 2} \right )\left ( \sqrt{5x + 7} \right )}$

$= \frac{15x – 10 – 15x – 21}{2 \left (3x – 2 \right )^{\frac{3}{2}}\left ( 5x + 7 \right )^{\frac{1}{2}}}$

$= \frac{-31}{2 \left (3x – 2 \right )^{\frac{3}{2}}\left ( 5x + 7 \right )^{\frac{1}{2}}}$

Example 3:

Find the derivative of $\frac{(x+3)^{4}}{\sqrt{x^{2}+5}}$

Solution:

Applying quotient rule, we have

$\frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{(x+3)^{4}}{\sqrt{x^{2}+5}} \right ) = \frac{\sqrt{x^{2}+5}.\frac{\mathrm{d} }{\mathrm{d} x}(x+3)^{4} – (x+3)^{4} . \frac{\mathrm{d} }{\mathrm{d} x} \left (\sqrt{x^{2}+5} \right )}{x^{2}+5}$

$= \frac{\sqrt{x^{2}+5}.4(x+3)^{3} – (x+3)^{4} . \frac{2x}{2\sqrt{x^{2}+5}} }{x^{2}+5}$

$= \frac{4. \left (x^{2}+5 \right ).(x+3)^{3} – x.(x+3)^{4} }{\left (x^{2}+5 \right )^{\frac{3}{2}}}$

$= \frac{\left ( x+3 \right )^{3}\left [ 4. \left (x^{2}+5 \right ) – x.(x+3) \right ]}{\left (x^{2}+5 \right )^{\frac{3}{2}}}$

$= \frac{\left ( x+3 \right )^{3}\left [ 4x^{2} + 20 – x^{2} – 3x \right ]}{\left (x^{2}+5 \right )^{\frac{3}{2}}}$

$= \frac{\left ( x+3 \right )^{3}\left [ 3x^{2} -3x + 20 \right ]}{\left (x^{2}+5 \right )^{\frac{3}{2}}}$

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