Question

Differentiate x^sinx+(sinx)^cosx with respect to x.

Answer 1

`y=x^(sinx)+(sinx)^cosx`

`=>y=e^(sinxlogx)+e^(cosxlogsinx)`

Differentiating both sides with respect to x, we get

`dy/dx=e^(sinx logx)xx d/dx(sinxlogx)+e^(cosxlog sinx)xxd/dx(cosx log sinx)`

`dy/dx=x^(sinx)(sinxxx1/x+logx xx cosx)+(sin)^(cosx)(cosx xx 1/sinx xx cosx +log sinx)`

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