jee-main 2017 Q79

jee-main · India · 02apr Differentiating Transcendental Functions Compute derivative of transcendental function

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

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

Paper Questions

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