JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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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{x} \cdot g(x)

, then

g(x)

equals

View answer and explanation

Correct option: B

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

Option B (9/(1+9x³)) is correct because after simplifying the inverse tangent expression and differentiating, the resulting g(x) is 9/(1+9x³).

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x

y

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\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

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v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}