Given: x^{sin x} + (sin x)^{cos x}

Let y= x^{sin x} + (sin x)^{cos x}

Let y = u + v

⇒ u = x^{sin x} and v = (sin x)^{cos x}

For, u = x^{sin x}

Taking log on both sides, we get

log u =log (x^{sin x} )

⇒log u = sin x.log(x)

Now, differentiate both sides with respect to x

For, v = (sin x)^{cos x}

Taking log on both sides, we get

log v =log(sin x)^{cos x}

⇒log v = cos x. log (sin x)

Now, differentiate both sides with respect to x

Because, y = u + v