Question

What is the derivative of `xe^(y^x)`?

Answer 1

`y'=(y/x+y^(x+1)lny)/(xy^x)`

Explanation:

`y=xe^(y^x)`

by taking the natural logarithm to both sides

`lny=lnxe^(y^x)`

using the properties of logarithmic functions :
`lnab=lna+lnb`

so,
`lny=lnx+lne^(y^x)`

`lne^(y^x)=y^xlne`=`y^x`
because `rarr`lne=1#

`lny=lnx+y^x` `rarr` `(1)`

let `u=y^x`

by taking the natural logarithm to both sides
`lnu=xlny`

Apply implicit differentiation

`(u')/u=(xy')/y+lny`

substitute for `u`

`u'=y^x(xy')/y+y^xlny`

now back to our original equation (1)

`lny=lnx+u`

Differentiate with respect to `x`
`(y')/y=1/x+u'`

substitute for `u'`

`y'=y/x+y(y^x)(xy')/y+yy^xlny`

simplify

`y'=(y/x+y^(x+1)lny)/(xy^x)`

I hope this was helpful.