If $y = {x^{{x^2}}},$ then $\frac{{dy}}{{dx}}$ equals

$(\log x + 1).\,{x^{{x^2} + 1}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y = {x^{{x^2}}},

then

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

equals

View answer and explanation

Correct option: D

{x^{{x^2} + 1}}.\,\,(\log (e{x^2}))

Option D (

{x^{{x^2} + 1}}.{\left( {\log (e{x^2})} \right)}

) is correct because it is the derivative of the function y = x^(x²) calculated using logarithmic differentiation.

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