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

Question

1542 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-19T15:10:18

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

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

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

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