If (y=x^{x^x}), then y' = | Mathematics | SolveeIt | Solveeit

Today, August 18

Question

If $y=x^{x^x}$ , then y' =

A

$\frac{-y^2}{x(1-y\log x)}$

B

$\frac{y^2}{1-y\log x}$

C

$\frac{y^2}{x(1-y\log x)}$

D

$\frac{-y^2}{1-y\log x}$

Answer

$\frac{y^2}{x(1-y\log x)}$

Explanation

Solution

The correct option is (C) : $\frac{y^2}{x(1-y\log x)}$ .