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

Quotient Rule of derivatives

{\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

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Azhar Ali

Azhar Ali

I graduated in Mathematics from the University of Sargodha, having master degree in Mathematics.

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