Offline paper
Mathematics · 2017
JEE Main 2017 · 2017-04-02 Offline · Q71
If for x ∈(0, 1/4), the derivative of tan^-1((6 x √ x)/(1-9 x^3)) is √ x · g(x), then g(x) equals:
If for $\displaystyle x \in\left(0, \frac{1}{4}\right)$, the derivative of $\displaystyle \tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)$ is $\displaystyle \sqrt{x} \cdot g(x)$, then $\displaystyle g(x)$ equals :
Official answer
(4)
$\displaystyle \frac{9}{1+9 x^3}$
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JEE Main 2017 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.