Question

y=cos(x^x) differentiate

Answer 1

Answer:

$\\\tt -{x}^{x} \sin( {x}^{x} ) (1 + logx)$

Step-by-step explanation:

Given:

$\\\tt y = \cos( {x}^{x} )$

Using chain rule :

$\\\tt \dfrac{dy}{dx} = (\dfrac{d \cos( {x}^{x} ) }{d {x}^{x} } ) .( \dfrac{d {x}^{x} }{dx} )$

The first partcan be done in one step for seng we'd have to use log.

Let ,

$\\\tt t = {x}^{x} \\\tt log(t) = x log(x)$

Differentiate both sides with respect to x,

$\\\tt \dfrac{dt}{dx} . \dfrac{1}{t} = x.( \dfrac{1}{x} ) + log(x)$

$\\\tt \dfrac{dt}{dx} = {x}^{x} (1 + logx)$

dt/dx is same as differentiation of x^x.

Now we can substitute this in the result we got after chain rule,

$\\\tt - {x}^{x} \sin( {x}^{x} ) (1 + logx)$