Question

Differentiate (sinx)^x + sin^-1√x  with respect to x.

Answer 1

Let y = (sinx)^x + sin^-1√x

Let u = (sinx)^x, v = sin^-1√x then we have y = u + v

Therefore,  dy/dx = du/dx + dv/dx ......(1)

Now, u = (sin x)^x 

Taking logarithm on both sides, we have log u = xlog(sinx). 

Differentiating both sides w.r.t. x, we have

Differentiating both sides w.r.t. x, we have