Differentiate the following w.r.t. x:
`x^(sinx)+(sinx)^(cosx)`

Explanation

"Let `u=x^ sinx` and `v=sinx ^cosx`
⇒`y=u+v`
diff on both side w.r.t. x
`dy/dx = (du)/dx +( dv)/dx`
Consider `u=x ^sinx`
taking log on bot side `logu=sinxlogx`
`1/(u) . (du)/dx` = `sinx/x + logx(cosx)`
`(du)/dx = u( sinx/x + logx(cosx))`
`(du)/dx` = `x^sinx( sinx/x + logx(cosx))`

Consider `v=(sinx)^cosx`
taking log on both side
`logv=cosxlogsinx`
`1/(v) . (dv)/dx` = `cosx/ (sinx) ( cosx) + log sinx( -sinx)`
`(dv)/dx = (sinx)^cosx ( cos^2(x)/sinx - sinx logsinx)`

hence =>
`dy/dx` = `x^sinx( sinx/x + logx(cosx))` + `sinx^cosx ( cos^2(x)/sinx - sinx logsinx)`