If y=\tan ^{-1}({3 x-x^{3}}/{1-3 x^{3}}) {-1}/{\\sqrt{3} \\leq x \\leq {1}/{\\sqrt{3} then \frac{d y}{d x} is

NIMCET 2021 Question Paper with solutions

Question : 46 of 120

Marks: +1, -0

y=\tan ^{-1}({3 x-x^{3}}/{1-3 x^{3}}) {-1}/{\√{3}} \≤ x \≤ {1}/{\√{3}}

then

{d y}/{d x}

Solution:

Given Question is

y=\tan ^{-1}({3 x-x^{3}}/{1-3 x^{3}})

Which is incorrect
Correct Question is :

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

y=3 \tan ^{-1} x

{d y}/{d x}={3}/{1+x^{2}}

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