Quotient Rule of Derivatives, Examples with Solutions -

Quotient Rule of Derivatives of der Here we will discuss Quotient Rule of derivatives in easy way if we have two functions $f(x)$ and $g(x)$ and if $f(x)$ and $g(x)$ are differentiable at $x$ and $g(x)$ is not equal to zero for any $x∈Dg$ then $f/g$ is differentiable at x and we will prove that

${\frac d{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\left[{\displaystyle\frac d{dx}}f\left(x\right)\right]g\left(x\right)-f\left(x\right)\left[{\displaystyle\frac d{dx}}g\left(x\right)\right]}{\left[g\left(x\right)\right]^2}}$ .

                            Suppose

$\phi\left(x\right)=\left[\frac{f\left(x\right)}{g\left(x\right)}\right]..........(1)$ .

$\phi\left(x+\delta x\right)=\left[\frac{f\left(x+\delta x\right)}{g\left(x+\delta x\right)}\right].........(2)$ .

$\phi\left(x+\delta x\right)-\phi\left(x\right)=\frac{f\left(x+\delta x\right)}{g\left(x+\delta x\right)}-\frac{f\left(x\right)}{g\left(x\right)}$ .

$\phi\left(x+\delta x\right)-\phi\left(x\right)=\frac{f\left(x+\delta x\right)g\left(x\right)-f\left(x\right)g\left(x+\delta x\right)}{g\left(x\right)g\left(x+\delta x\right)}........(3)$ .

             Subtracting and Adding  in numerator and  in .

$\begin{array}{l}\phi\left(x+\delta x\right)-\phi\left(x\right)=\frac{f\left(x+\delta x\right)g\left(x\right)-f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x+\delta x\right)+f\left(x\right)g\left(x\right)}{g\left(x\right)g\left(x+\delta x\right)}\end{array}$ .

${\phi\left(x+\delta x\right)-\phi\left(x\right)=\frac1{f\left(x\right)g\left(x+\delta x\right)}\left[f\left(x+\delta x\right)g\left(x\right)-f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x+\delta x\right)+f\left(x\right)g\left(x\right)\right]}$ .

$⇒\;\;\;\;\;\;\;\frac{\phi\left(x+\delta x\right)-\phi\left(x\right)}{\delta x}=\frac1{f\left(x\right)g\left(x+\delta x\right)}\left[\frac{f\left(x+\delta x\right)-f\left(x\right)}{\delta x}g\left(x\right)-f\left(x\right)\frac{g\left(x+\delta x\right)-g\left(x\right)}{\delta x}\right]$ .

$⇒\;\;\;\;\;\;\phi'\left(x\right)=\frac1{g\left(x\right)g\left(x\right)}\left[f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)\right]$ .

Which is required quotient rule

$⇒\;\;\;\;\;\;\boxed{\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\displaystyle\frac d{dx}\left[f\left(x\right)\right]g\left(x\right)-f\left(x\right)\frac d{dx}\left[g\left(x\right)\right]}{\displaystyle\left[g\left(x\right)\right]^2}}$ .

Example 1: [Quotient Rule of Derivatives]

let $f(x)=3x$ and $g(x)=2x^2$

Now we use quotient rule

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac d{dx}\left(\frac{3x}{2x^2}\right)$ .

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{2x^2{\displaystyle\frac d{dx}}\left(3x\right)-3x{\displaystyle\frac d{dx}}2x^2}{\left(2x^2\right)^2}$ .

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{2x^2\left(3\right){\displaystyle\frac{dx}{dx}}-3x\left(2\right){\displaystyle\frac d{dx}}x^2}{\left(2x^2\right)^2}$

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{2x^2\left(3\right)\left(1\right)-3x\left(2\right)\left(2x\right)}{\left(2x^2\right)^2}$

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{6x^2-12x^2}{2^2x^4}$

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{-6x^2}{4x^4}$

$\boxed{\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{-3x}{2x^2}}$

Example 2: [Quotient Rule of derivatives]

Let $f(x)=4x$ and $g(x)=9x^4$

Now we use Quotient Rule

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac d{dx}\left(\frac{4x}{9x^4}\right)$

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\displaystyle9x^4\frac d{dx}\left(4x\right)-\left(4x\right)\frac d{dx}\left(9x^2\right)}{\displaystyle\left(9x^4\right)^2}$

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\displaystyle9x^4\left(4\right)\frac d{dx}\left(x\right)-\left(4x\right)\left(9\right)\frac d{dx}\left(x^2\right)}{\displaystyle\left(9\right)^2\left(x^4\right)^2}$

$\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\displaystyle9x^4\left(4\right)\left(1\right)-\left(4x\right)\left(9\right)\left(2x\right)}{\displaystyle\left(9\right)^2\left(x^4\right)^2}$