Question

Differentiate \[{{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)\] with respect to \[\sqrt{1+4{{x}^{2}}}\].

Answer 1

Hint: We use the Quotient rule for Derivative of a function\[f(x)\] with respect to another function \[g(x)\] is given as \[\dfrac{d\left( f\left( x \right) \right)}{d\left( g\left( x \right) \right)}=\dfrac{d(f(x))}{dx}\div \dfrac{d\left( g\left( x \right) \right)}{dx}\].

Complete step by step answer:
Here, we are asked to differentiate a function, say \[f\left( x \right)={{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)\] with respect to \[g\left( x \right)=\sqrt{1+4{{x}^{2}}}\].
i.e. we need to determine \[\dfrac{d\left( f\left( x \right) \right)}{d\left( g\left( x \right) \right)}\].
We know, \[\dfrac{d\left( f\left( x \right) \right)}{d\left( g\left( x \right) \right)}=\dfrac{d}{dx}f(x)\times \dfrac{dx}{d(g(x))}...........\]equation\[(1)\]
Now, first we will find the value of the derivative of \[f\left( x \right)\].
We are given \[f\left( x \right)={{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)\]
So, \[\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}{{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)\]
We can see that \[f(x)\] is a composite function , i.e it is of the type \[f(x)=h(p(x))\] , where \[h(x)={{\tan }^{-1}}(x)\] and \[p(x)=\dfrac{1+2x}{1-2x}\] . So , the derivative of \[f(x)\] is given as \[f'(x)=h'(p(x))\times p'(x)\]
Now , we know, \[\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( x \right) \right)=\dfrac{1}{1+{{x}^{2}}}\]
So, \[\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right) \right)=\dfrac{1}{1+{{\left( \dfrac{1+2x}{1-2x} \right)}^{2}}}.\dfrac{d}{dx}\left( \dfrac{1+2x}{1-2x} \right)........\]equation\[(2)\]
Now, first we will determine the value of the derivative of \[\dfrac{1+2x}{1-2x}\].
To find the value of the derivative of \[\dfrac{1+2x}{1-2x}\], we need to apply quotient rule. The quotient rule of differentiation is given as \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v{{u}^{'}}-u{{v}^{'}}}{{{v}^{2}}}\].
Now , applying quotient rule to find the derivative , we get ,
\[\dfrac{d}{dx}\left( \dfrac{1+2x}{1-2x} \right)=\dfrac{\left( 1-2x \right).2-\left( -2 \right).\left( 1+2x \right)}{{{\left( 1-2x \right)}^{2}}}\]
\[=\dfrac{2-4x+2+4x}{{{\left( 1-2x \right)}^{2}}}\]
\[=\dfrac{4}{{{\left( 1-2x \right)}^{2}}}\]
Now , we will substitute the value of \[\dfrac{d}{dx}\left( \dfrac{1+2x}{1-2x} \right)\] in equation\[(2)\].
On substituting the value of \[\dfrac{d}{dx}\left( \dfrac{1+2x}{1-2x} \right)\] in equation\[(2)\], we get \[\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right) \right)=\dfrac{1}{1+{{\left( \dfrac{1+2x}{1-2x} \right)}^{2}}}.\dfrac{4}{{{\left( 1-2x \right)}^{2}}}\]
\[=\dfrac{1}{\dfrac{{{\left( 1-2x \right)}^{2}}+{{\left( 1+2x \right)}^{2}}}{{{\left( 1-2x \right)}^{2}}}}.\dfrac{4}{{{\left( 1-2x \right)}^{2}}}\]
\[=\dfrac{{{\left(1-2x\right)}^{2}}}{{{\left(1-2x\right)}^{2}}+{{\left(1+2x \right)}^{2}}}.\dfrac{4}{{{\left( 1-2x \right)}^{2}}}\]

\[=\dfrac{2}{1+4{{x}^{2}}}\]
So , \[\dfrac{d}{dx}f(x)=\dfrac{2}{1+4{{x}^{2}}}\]
Now , we will find the derivative of\[g(x)=\sqrt{1+4{{x}^{2}}}\]
.
So, \[\dfrac{d}{dx}g\left( x \right)=\dfrac{d}{dx}\sqrt{1+4{{x}^{2}}}\]
Again \[g(x)\] is a composite function. So,
\[\dfrac{d}{dx}g\left( x \right)=\dfrac{1}{2\sqrt{1+4{{x}^{2}}}}.\dfrac{d}{dx}(1+4{{x}^{2}})\]
\[=\dfrac{4x}{\sqrt{1+4{{x}^{2}}}}\]
Now, Now , we know inverse function theorem of differentiation says that if we are given \[g(x)\] and \[\dfrac{d}{dx}g(x)=g'(x)\] then , \[\dfrac{dx}{d(g(x))}=\dfrac{1}{g'(x)}\] .
So, \[\dfrac{dx}{d(g(x))}=\dfrac{1}{g'(x)}=\dfrac{1}{\dfrac{4x}{\sqrt{1+4{{x}^{2}}}}}=\dfrac{\sqrt{1+4{{x}^{2}}}}{4x}\]

Now, the derivative of \[{{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)\] with respect to \[\sqrt{1+4{{x}^{2}}}\]can be calculated by substituting the values of derivative of \[\dfrac{d}{dx}f(x)\] and \[\dfrac{dx}{d(g(x))}\] in equation\[(1)\].
\[\dfrac{d{{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)}{d\left( \sqrt{1+4{{x}^{2}}} \right)}=\dfrac{2}{1+4{{x}^{2}}}\times \dfrac{\sqrt{1+4{{x}^{2}}}}{4x}\]
\[=\dfrac{1}{2x\sqrt{1+4{{x}^{2}}}}\]
Hence , the derivative of \[{{\tan }^{-1}}\left( \dfrac{1+2x}{1-2x} \right)\] with respect to \[\sqrt{1+4{{x}^{2}}}\] is given as \[\dfrac{1}{2x\sqrt{1+4{{x}^{2}}}}\]
Note: While differentiating \[\sqrt{1+4{{x}^{2}}}\]with respect to \[x\], most students make a mistake of writing \[\dfrac{d}{dx}\left( \sqrt{1+4{{x}^{2}}} \right)=\dfrac{1}{2\sqrt{1+4{{x}^{2}}}}\] which is wrong.
\[\dfrac{d}{dx}\sqrt{1+4{{x}^{2}}}\] \[=\dfrac{1}{2\sqrt{1+4{{x}^{2}}}}\times 8x\]. Such mistakes should be avoided as students can end up getting a wrong answer due to such mistakes.