Quotient Rule

In Calculus, the Quotient rule is a method for determining the derivative (differentiation) of a function which is the ratio of two function that are differentiable in nature.

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:

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

It is said as Differentiation of 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.

Proof for Quotient Rule:

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:

$\large \left ( \frac{s}{t} \right )'(x) = \lim \limits_{h \to 0} \frac{\left (\frac{s}{t} \right )(x+h)- \left ( \frac{s}{t} \right )(x)}{h}$

$\large = \lim \limits_{h \to 0} \frac{\left (\frac{s(x+h)}{t(x+h)} \right )- \left ( \frac{s(x)}{t(x)} \right )}{h}$

$\large = \lim \limits_{h \to 0} \frac{\left (\frac{s(x+h).t(x) – s(x).t(x+h)}{t(x+h).t(x)} \right )}{h}$

$\large = \lim \limits_{h \to 0} \frac{1}{h}\left (\frac{s(x+h).t(x) – s(x).t(x+h)}{t(x+h).t(x)} \right )$

Adding and subtracting s(x).t(x) in the numerator, we have

$\large = \lim \limits_{h \to 0} \frac{1}{h}\left (\frac{s(x+h).t(x) – s(x).t(x+h) + s(x).t(x) – s(x).t(x)}{t(x+h).t(x)} \right )$

$\large = \lim \limits_{h \to 0} \frac{1}{h}\left (\frac{s(x+h).t(x) – s(x).t(x)}{t(x+h).t(x)} \right ) – \lim \limits_{h \to 0} \frac{1}{h}\left (\frac{s(x).t(x+h) – s(x).t(x)}{t(x+h).t(x)} \right )$

$\large = \lim \limits_{h \to 0} \frac{1}{h}\left (\frac{s(x+h) – s(x)}{t(x+h)} \right ) – \lim \limits_{h \to 0} \frac{1}{h} \left (\frac{\left (t(x+h) – t(x) \right ).s(x)}{t(x+h).t(x)} \right )$

$\large = \frac{1}{t(x)} \lim \limits_{h \to 0} \left (\frac{s(x+h) – s(x)}{h} \right ) – \left (\frac{s(x)}{t(x).t(x)} \right )\lim \limits_{h \to 0} \left (\frac{\left (t(x+h) – t(x) \right )}{h} \right )$

$\large = \frac{1}{t(x)} \left ( s'(x) \right ) – \left (\frac{s(x)}{\left (t(x) \right )^{2}} \right ).t'(x)$

$\large \left ( \frac{s}{t} \right )'(x) = \frac{s'(x).t(x) – t'(x).s(x)}{\left (t(x) \right )^{2}}$

Let us work out some examples:

Examples: 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$

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

Solution: Applying quotient rule, we have

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

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

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

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

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

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