Question

Differentiate the function with respect to x.

`x^x - 2^(sin x)`

Answer 1

Let, y = x^x - 2^{sin x}

Again, let u = x^x, v = 2^{sin x}

y = u - v
Taking logarithm of both sides of u = x^x,

log u = log x^x = x log x

Differentiating both sides with respect to x,

`1/u (du)/dx = x d/dx log x + log x d/dx (x)`

`=> 1/u (du)/dx = x * 1/x + log x xx 1/u (du)/dx = 1 + log x`       ...(1)

`therefore (du)/dx = u (1 + log x) = x^x (1 + log x)`        ...(2)

Now, from `v = 2^(sin x)`

`(dv)/dx= 2^ (sin x) log  2  d/dx (sin x)`

`= 2^(sin x) log  2  cos x`        ...(3)

From equation (1), y = u – v

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

Putting the values ​​of `(du)/dx` from equation (2) and `(dv)/dx` from (3),

`dy/dx = x^x (1 + log x) - 2^(sin x) (cos x. log  2)`