Question

Given that y = (sin x)^x . x^sinx + a^x, find `dy/dx`.

Answer 1

y = (sin x)^x .  x^sinx + a^x

Let u = (sin x)^x; v = x^sinx

y = u . v + a^x   ...(i)

`dy/dx = u(dv)/dx + v (du)/dx + a^xloga`

u = (sin x)^x

Taking log on both sides

log u = x log sin x

Differentiate log u = `1/u (du)/dx = x/sinx xx cosx + log sinx`

`= (du)/dx = (sin)^x[x cot x + logsinx]`

v = x^sinx   ...(ii)

log v = sin x log x

`1/v (dv)/dx = sinx/x + logx cosx`

`(dv)/dx = x^sinx [sinx/x + log cos x]`   ...(iii)

Put eq (ii) and (iii) in eq (i)

`dy/dx = (sinx)^x .x^sinx [sinx/x+log cos x] + x^sinx.(sinx)^x[x cotx+log sinx] + a^x loga`