JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q71

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MathematicsSingle correct+4 / -1Question 71 of 90

If for $x\in\left(0,\dfrac14\right)$ , the derivative of $\tan^{-1}\left(\dfrac{6x\sqrt x}{1-9x^3}\right)$ is $\sqrt x\cdot g(x)$ , then $g(x)$ equals :

\dfrac{3x\sqrt x}{1-9x^3}

\dfrac{3x}{1-9x^3}

\dfrac3{1+9x^3}

\dfrac9{1+9x^3}

Answer and worked solution

Official answer: D

With $t=3x^{3/2}$ (and $t<1$ on this interval), the function is $2\tan^{-1}t$ . Its derivative is
$\frac2{1+9x^3}\cdot\frac92x^{1/2}=\sqrt x\cdot\frac9{1+9x^3}.$

Topics:Chain rule and implicit differentiation Inverse trigonometric functions

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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 71