Question

Differentiate the function with respect to x.

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

Answer 1

Let, x^{sin x} + (sin x)^{cos x}

Again, let y = u + v

Differentiating both sides with respect to x,

`(dy)/dx = (du)/dx + (dv)/dx`    ...(1)

अब, u = x^{sin x}

Taking logarithm of both sides,

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

On differentiating both sides with respect to,

`1/u du/dx = sin x d/dx log x + log x d/dx sin x`

 = `sin x . 1/x + log x * cos x = cos x log x + sin x/x`

`therefore (du)/dx = u (cos x log x + (sin x)/x) = x^(sin x) (cos x log x + (sin x)/x)`   ....(2)

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

Taking logarithm of both sides,

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

On differentiating both sides with respect to,

`1/v (dv)/dx = cos x  d/dx log sin x + log sin x  d/dx  cos x`

`= cos x * 1/(sin x) d/dx  sin x + log sin x * (- sin x)`

`= cos x * 1/sin x * cos x - sin x log sin x`

`= - sin x log sin x + cot x * cos x`

`therefore dv/dx = v [-sin x log sin x + cot x cos x]`

`= (sin x)^(cos x) [-sin x log sin x + cot x cos x]`          ....(3)

Putting the values ​​of `(du)/dx` and `(dv)/dx` from equations (2) and (3) in equation (1), we get,

`therefore dy/dx = (du)/dx + (dv)/dx`

`= x^(sin x) (cos x log x + sin x/x) + (sin x)^(cos x) [- sin x log sin x +cot x cos x]`