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).
eq(2)-eq(1).
\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).
| \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}. |
Subtracting and Adding f(x)g(x) in numerator and in eq (3).
\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}
\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\displaystyle36x^4-72x^2}{\displaystyle81x^8}
\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{36x^4}{81x^8}-\frac{72x^2}{81x^8}
\boxed{\frac d{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{12}{27x^4}-\frac{24}{27x^6}}
you can also see product rule