Question

What is the derivative of `y=x^2+x^x`?

Answer 1

`2x + x^x (1+lnx)`

Explanation:

The required derivative would be the sum of the derivative of `x^2` and `x^x`.

The derivative of `x^2` is 2x. The derivative of `x^x` can arrived at as follows:

Let `f=x^x`. Tale log on both sides, ln f= x ln x. Now differentiate both sides, `(1/f) (df)/dx = ln x +x(1/x)` (product rule).

Thus `(df)/dx= x^x (1+lnx)`.

`(dy)/dx= 2x + x^x (1+lnx)`