Differentiate the following w.r.t. x: `(xcosx)^x+(xsinx)^(1/x)`

Explanation

Let `y = (xcosx)^x+(xsinx)^(1/x)`
Let `u = (xcosx)^x and v = (xsinx)^(1/x)`
Then, `y = u+v`
`dy/dx = (du)/dx+(dv)/dx`
Now, `u = (xcosx)^x`
Taking log both sides,
`log u = x(logx+logcosx)`
Differentiating both sides,
`1/u (du)/dx = (logx+cosx)*1+x(1/x+1/cosx(-sinx))`
`1/u (du)/dx = (logx+cosx)+1-xtanx`
`(du)/dx = (xcosx)^x(log(xcosx)+1-xtanx)->(1)`
Now, `v = (xsinx)^(1/x)`
Taking log both sides,
`log v = 1/x(logx+logsinx)`
Differentiating both sides,
`1/v (dv)/dx = 1/x^2(x(1/x+1/sinx(cosx))-(logx+logsinx)*1)`
`1/v(dv)/dx = 1/x^2(1+cotx -log(xsinx))`
`(dv)/dx = (xsinx)^(1/x)1/x^2(1+cotx -log(xsinx))`
So, `dy/dx = (du)/dx+(dv)/dx`
`dy/dx = (xcosx)^x(log(xcosx)+1-xtanx)+(xsinx)^(1/x)((1+cotx -log(xsinx))/x^2)`