find dy/dx when x^sinx + (sinx)^cosx

y= x^sinx + (sinx)^cosx
​let x^sinx = u and (sinx)^cosx= v
dy/dx = du/dx + dv/dx
u= x^sinx+++
Taking log on both sides
log u = log(x^sinx)
       = sinx log x
Differntiating both sides
1/u du/dx = cosx logx + sinx/x (product rule)
du/dx= u {cosxlogx+sinx/x}
du/dx= x^sinx(cosxlogx+sinx/x)

Now, v = (sinx)^cosx
Taking log on both sides
log v = log {(sinx)^cosx}
        = cosx log sinx
Differentiating
1/v. dv/dx = (-sinx)log sinx + cos x (cos x/ sin x)
dv/dx = v { cos x cot x - sinx log sinx}
​dv/dx = (sinx)^cos x (cos x cot x - sinx log sinx)
 
=> dy/dx = { x^sinx(cosxlogx+sinx/x)} + { ​ (sinx)^cos x (cos x cot x - sinx log sinx)}
Therefore, dy/dx = x^sinx (cosxlog x + sinx/x) + sinx^cosx (cosx cotx - sinx log sinx)

Note- I've solved both the terms separately because my teacher says log on a whole is not applicable on addition and subtraction operations.
Hope this helps ! :-)