Question

find the derivative of x^x + (log x)^x​

Answer 1

Answer:

Step-by-step explanation:

We have,

$y=x^{x} + \{log(x)\}^{x}$

Differentiating both sides w.r.t x, we have,

$\frac{dy}{dx}={x}^{x}\[\frac{d}{dx}\{x.log(x)\}\] + \{log(x)\}^{x}\[\frac{d}{dx}\{x.log(log(x))\}\]$

$\frac{dy}{dx}={x}^{x}\{1+log(x)\} + \{log(x)\}^{x}\{log(log(x))+\frac{1}{log(x)}\}$