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

⇒ y = a + b

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

a= x^{cos x}

Taking log both the sides:

⇒ log a= log (x)^{cos x}

⇒ log a= cos x log x

{log x^a = alog x}

Differentiating with respect to x:

b = (sin x)^{tan x}

Taking log both the sides:

⇒ log b= log (sin x)^{tan x}

⇒ log b= tan x log (sin x)

{log x^a = alog x}

Differentiating with respect to x: