# Quotient Rule of Derivatives

The quotient rule of derivatives is $\frac{d}{{dx}}\left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{g\left( x \right)f’\left( x \right) – f\left( x \right)g’\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$. In other words, we can read this as the derivative of a quotient of two functions is equal to the second function as it is and the derivative of the first function minus the first function as it is and the derivative of the second function divided by the square of the second function. This rule can be proved using the first principle or derivative by definition.

Consider a function of the form $y = \frac{{f\left( x \right)}}{{g\left( x \right)}}$.

First we take the increment or small change in the function:
$\begin{gathered}y + \Delta y = \frac{{f\left( {x + \Delta x} \right)}}{{g\left( {x + \Delta x} \right)}} \\ \Rightarrow \Delta y = \frac{{f\left( {x + \Delta x} \right)}}{{g\left( {x + \Delta x} \right)}} – y \\ \end{gathered}$

Putting the value of function $y = \frac{{f\left( x \right)}}{{g\left( x \right)}}$ in the above equation, we get
$\begin{gathered} \Rightarrow \Delta y = \frac{{f\left( {x + \Delta x} \right)}}{{g\left( {x + \Delta x} \right)}} – \frac{{f\left( x \right)}}{{g\left( x \right)}} \\ \Rightarrow \Delta y = \frac{{f\left( {x + \Delta x} \right)g\left( x \right) – f\left( x \right)g\left( {x + \Delta x} \right)}}{{g\left( {x + \Delta x} \right)g\left( x \right)}} \\ \end{gathered}$

Subtracting and adding $f\left( x \right)g\left( x \right)$ on the right hand side, we have
$\begin{gathered}\Rightarrow \Delta y = \frac{{f\left( {x + \Delta x} \right)g\left( x \right) – f\left( x \right)g\left( x \right) + f\left( x \right)g\left( x \right) – f\left( x \right)g\left( {x + \Delta x} \right)}}{{g\left( {x + \Delta x} \right)g\left( x \right)}} \\ \Rightarrow \Delta y = \frac{{g\left( x \right)\left[ {f\left( {x + \Delta x} \right) – f\left( x \right)} \right] + f\left( x \right)\left[ {g\left( x \right) – g\left( {x + \Delta x} \right)} \right]}}{{g\left( {x + \Delta x} \right)g\left( x \right)}} \\ \Rightarrow \Delta y = \frac{{g\left( x \right)\left[ {f\left( {x + \Delta x} \right) – f\left( x \right)} \right] – f\left( x \right)\left[ {g\left( {x + \Delta x} \right) – g\left( x \right)} \right]}}{{g\left( {x + \Delta x} \right)g\left( x \right)}} \\ \end{gathered}$

Dividing both sides by $\Delta x$, we get
$\begin{gathered}\Rightarrow \frac{{\Delta y}}{{\Delta x}} = \frac{{g\left( x \right)\left[ {f\left( {x + \Delta x} \right) – f\left( x \right)} \right] – f\left( x \right)\left[ {g\left( {x + \Delta x} \right) – g\left( x \right)} \right]}}{{\Delta xg\left( {x + \Delta x} \right)g\left( x \right)}} \\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = \left[ {\frac{{f\left( {x + \Delta x} \right) – f\left( x \right)}}{{\Delta x}}} \right]\frac{{g\left( x \right)}}{{g\left( {x + \Delta x} \right)g\left( x \right)}} – \frac{{f\left( x \right)}}{{g\left( {x + \Delta x} \right)g\left( x \right)}}\left[ {\frac{{g\left( {x + \Delta x} \right) – g\left( x \right)}}{{\Delta x}}} \right] \\ \end{gathered}$

Taking the limit of both sides as $\Delta x \to 0$, we have
$\begin{gathered}\Rightarrow \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \left[ {\frac{{f\left( {x + \Delta x} \right) – f\left( x \right)}}{{\Delta x}}} \right]\mathop {\lim }\limits_{\Delta x \to 0} \frac{{g\left( x \right)}}{{g\left( {x + \Delta x} \right)g\left( x \right)}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, – \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( x \right)}}{{g\left( {x + \Delta x} \right)g\left( x \right)}}\mathop {\lim }\limits_{\Delta x \to 0} \left[ {\frac{{g\left( {x + \Delta x} \right) – g\left( x \right)}}{{\Delta x}}} \right] \\ \Rightarrow \frac{{dy}}{{dx}} = f’\left( x \right)\frac{{g\left( x \right)}}{{g\left( {x + 0} \right)g\left( x \right)}} – \frac{{f\left( x \right)}}{{g\left( {x + 0} \right)g\left( x \right)}}g’\left( x \right) \\ \frac{{dy}}{{dx}} = \frac{{g\left( x \right)f’\left( x \right) – f\left( x \right)g’\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}} \\ \frac{d}{{dx}}\left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{g\left( x \right)f’\left( x \right) – f\left( x \right)g’\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}} \\ \end{gathered}$

Example: Find the derivative of $y = \frac{{{x^3} – 8}}{{{x^3} + 8}}$

We have the given function as
$y = \frac{{{x^3} – 8}}{{{x^3} + 8}}$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{{{x^3} – 8}}{{{x^3} + 8}}} \right)$

Now using the quotient rule of a derivative, we have
$\begin{gathered}\frac{{dy}}{{dx}} = \frac{{\left( {{x^3} + 8} \right)\frac{d}{{dx}}\left( {{x^3} – 8} \right) – \left( {{x^3} – 8} \right)\frac{d}{{dx}}\left( {{x^3} + 8} \right)}}{{{{\left( {{x^3} + 8} \right)}^2}}} \\ \frac{{dy}}{{dx}} = \frac{{\left( {{x^3} + 8} \right)3{x^2} – \left( {{x^3} – 8} \right)3{x^2}}}{{{{\left( {{x^3} + 8} \right)}^2}}} \\ \frac{{dy}}{{dx}} = \frac{{\left( {3{x^5} + 24{x^2}} \right) – \left( {3{x^5} – 24{x^2}} \right)}}{{{{\left( {{x^3} + 8} \right)}^2}}} \\ \frac{{dy}}{{dx}} = \frac{{48{x^2}}}{{{{\left( {{x^3} + 8} \right)}^2}}} \\ \end{gathered}$