Xsinx +(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 ! :-)