Find , when

y = x^x + (sin x)^x

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

⇒ y = a + b

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

a= x^x

Taking log both the sides:

⇒ log a= log (x)^x

⇒ log a= x log x

{log x^a = alog x}

Differentiating with respect to 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: