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

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

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

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

If for

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

, the derivative of

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

\sqrt{\mathrm{x}} \cdot \mathrm{g}(\mathrm{x})

, then

g(x)

equals :

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the expression for g(x) after simplifying and differentiating the given function.