if y= x^x + (sinx)^x then find dy/dx
plzz help

Differentiate part by part let u=x^x  nd v=sin^x        

y=u+v

u=x^x  take log both sides log u= x. logx Now start differentiating  1/u.(du/dx)=1+logx  or du/dx=u.(1+log x) now put u value we will get x^x differentiation.

v=sinx^x take log both sides log v=x.log sinx or  1/v .(dv/dx)=x.(sinx)^-1 cosx+log sinx  Similarly we have found dv/dx . 

Add both d solution du/dx+dv/dx thats d answer of dy/dx