Skip to main content
Back to DIFFERENTIATION

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Hard
1 of 1
Analyzing +4 | -1
The expression of dydx\frac{{dy}}{{dx}} of the function y=axax…∞y = {a^{{x^a}{{^x}^{ \ldots \infty }}}}, is
View answer and explanation
Correct option: C
y2log⁡yx(1−ylog⁡xlog⁡y)\frac{{{y^2}\log y}}{{x\left( {1 - y\log x\log y} \right)}}
The correct answer is C (y2log⁡yx(1−ylog⁡xlog⁡y)\frac{{{y^2}\log y}}{{x\left( {1 - y\log x\log y} \right)}}) because it is the result of logarithmic differentiation and solving for dydx\frac{dy}{dx}.

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.