Find dy/dx, when y = x^(sin x) + (sin x)^x

Question

Find \(\cfrac{dy}{d\text x}\), when

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

Answer 1

Let y = x^{sin x} + (sin x)^x

⇒ y = a + b

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

  \(\Bigg\{\) Using chain rule, \(\cfrac{d(u +a)}{d\text x}=\cfrac{du}{d\text x}+\cfrac{da}{d\text x}\) where a and u are any variables \(\Bigg\}\)

a= x sin^x

Taking log both the sides:

⇒ log a= log x^{sin x}

⇒ log a= sin x log x

{log x^a = a log 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 = a log x}

Differentiating with respect to x:

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