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

Hint: In these types of questions assume the first function ${\tan ^{ - 1}}(\dfrac{{1 + 2x}}{{1 - 2x}})$ be u and second function $\sqrt {1 + 4{x^2}} $be v. Now according to question, you need to find $\dfrac{{du}}{{dv}}$. Now$\dfrac{{du}}{{dv}}$ can also be written as $\dfrac{{\dfrac{{du}}{{\dfrac{{dx}}{{dv}}}}}}{{dx}}$. So, differentiate functions u and v separately and then find the value of $\dfrac{{du}}{{dv}}$ to obtain the solution to the problem.

Complete step-by-step answer:
Let ${\tan ^{ - 1}}(\dfrac{{1 + 2x}}{{1 - 2x}})$ be u and $\sqrt {1 + 4{x^2}} $be v.
Now, let’s find $\dfrac{{du}}{{dx}}$and $\dfrac{{dv}}{{dx}}$ separately.
We know that, any inverse function of the form ${\tan ^{ - 1}}(\dfrac{{A + B}}{{A - B}}) = {\tan ^{ - 1}}A + {\tan ^{ - 1}}B$.
Now, in this case ${\tan ^{ - 1}}(\dfrac{{1 + 2x}}{{1 - 2x}})$ can be written as ${\tan ^{ - 1}}(1) + {\tan ^{ - 1}}(2x)$
So, $\dfrac{{du}}{{dx}}$ = $\dfrac{d}{{dx}}$(${\tan ^{ - 1}}(1) + {\tan ^{ - 1}}(2x)$)
Now we also know that ${\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}$
so, $\dfrac{{du}}{{dx}}$ = (0+$\dfrac{d}{{dx}}$$(\dfrac{1}{{1 + {x^2}}})$)
$\dfrac{{du}}{{dx}}$=$\dfrac{2}{{1 + 4{x^2}}}$ - equation (1)
So, we have the value of $\dfrac{{du}}{{dx}}$, now we need the value of \[\dfrac{{dv}}{{dx}}\]=$\dfrac{d}{{dx}}(\sqrt {1 + 4{x^2}} )$
now any function of the form $\dfrac{{d(\sqrt f )}}{{dx}} = \dfrac{1}{{2\sqrt f }}$
so, \[\dfrac{{dv}}{{dx}}\]=$\dfrac{1}{2} \times \dfrac{1}{{\sqrt {1 + 4{x^2}} }}(0 + 8x)$
\[\dfrac{{dv}}{{dx}}\]=$\dfrac{{4x}}{{\sqrt {1 + 4{x^2}} }}$ -equation (2)
Now to find \[\dfrac{{du}}{{dv}}\] we just need to divide equation (1) from equation (2)
\[\dfrac{{du}}{{dv}}\]=$\dfrac{2}{{1 + 4{x^2}}} \times \dfrac{{\sqrt {1 + 4{x^2}} }}{{4x}}$
\[\dfrac{{du}}{{dv}}\]=$\dfrac{1}{{2x\sqrt {1 + 4{x^2}} }}$


Note: In questions where you have to differentiate trigonometric functions, identities and formulas are always the primary key. It helps you to simplify the solution, for example in this question we used three different properties that are
1)$\dfrac{{d(\sqrt f )}}{{dx}} = \dfrac{1}{{2\sqrt f }}$
2)${\tan ^{ - 1}}(\dfrac{{A + B}}{{A - B}}) = {\tan ^{ - 1}}A + {\tan ^{ - 1}}B$
3)${\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}$
so, try to remember as many properties and identities as possible.