JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

y = {\tan ^{ - 1}}\left( {\frac{{\log \left( {\frac{e}{{{x^2}}}} \right)}}{{\log \left( {e{x^2}} \right)}}} \right) + {\tan ^{ - 1}}\left( {\frac{{3 + 2\log x}}{{1 - 6\log x}}} \right)

, then

\frac{{{d^2}y}}{{d{x^2}}}

View answer and explanation

Correct option: C

Option C (0) is correct because after simplifying the expression for y, we find that y is a constant, and therefore its second derivative is 0.

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