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 $\sqrt{x} \cdot \mathrm{~g}(x)$, then $\mathrm{g}(x)$ equals :

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

JEE Main · 2017 · 02 Apr 2017 · Offline · Differentiation

If for

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

, the derivative of

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

x \cdot g (x)

, then

g (x)

equals :

Answer: (B)

Substituting

t = x^{3/2}

turns the argument into

\frac{6 t}{1 - 9 t ^{2}} = \frac{2 ( 3 t )}{1 - ( 3 t ) ^{2}}

, the double-angle form, so the inverse tangent is

2 tan^{- 1} (3 t) = 2 tan^{- 1} (3 x^{3/2})

. Differentiating,

\frac{d y}{d x} = 2 \cdot \frac{1}{1 + 9 x ^{3}} \cdot \frac{9}{2} x^{1/2} = x \cdot \frac{9}{1 + 9 x ^{3}}

, so

g (x) = \frac{9}{1 + 9 x ^{3}}

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option