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let y = x^{sin x} + (sin x)^x

⇒ y = a + b

where a= x^{sin x}; b = (sin x)^x

a= x^{sin x}

Taking log both the sides:

⇒ log a= log x^{sin x}

⇒ log a= sin x log x

{log x^a = alog x}

Differentiating with respect to x:

Put the value of a = x^{sin x} :

b = (sin x)^x

Taking log both the sides:

⇒ log b= log (sin x)^x

⇒ log b= x log (sin x)

{log x^a = alog x}

Differentiating with respect to x:

Put the value of b = (sin x)^x :