If $y = {\log _2}{\log _2}\left( x \right)$, then $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ is equal to

$\frac{{{{\log }_2}e}}{{{{\log }_e}x}}$

$\frac{{{{\log }_2}e}}{{x{{\log }_x}2}}$

$\frac{{{{\log }_2}x}}{{{{\log }_e}2}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y = {\log _2}{\log _2}\left( x \right)

, then

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

is equal to

View answer and explanation

Correct option: D

\frac{{{{\log }_2}e}}{{x{{\log }_e}x}}

The correct answer is D (

\frac{{{{\log }_2}e}}{{x{{\log }_e}x}}

) because it is the result of correctly applying the chain rule and change of base formula for logarithms.

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