For $x \in \left( {0,\frac{1}{4}} \right),$ if the derivative of ${\tan ^{ - 1}}\left( {\frac{{6x\sqrt x }}{{1 - 9{x^3}}}} \right)$ is $\sqrt x \cdot g\left( x \right)$, then $g(x)$ equals

$\frac{{3x\sqrt x }}{{1 - 9{x^3}}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

For

x \in \left( {0,\frac{1}{4}} \right),

if the derivative of

{\tan ^{ - 1}}\left( {\frac{{6x\sqrt x }}{{1 - 9{x^3}}}} \right)

\sqrt x \cdot g\left( x \right)

, then

g(x)

equals

View answer and explanation

Correct option: A

\frac{9}{{1 + 9{x^3}}}

Option D (

\frac{3}{{1 + 9{x^3}}}

) is incorrect. Option A (

\frac{9}{{1 + 9{x^3}}}

) is correct because it is the derivative of the given function after simplification using the arctangent double angle formula.