1. ## Question x^x

If $\displaystyle \frac{{df}}{{dx}}=x^x$ for x>0, and $\displaystyle g(x)=f(x^2)$ then $\displaystyle \frac{{dg}}{{dx}} =$ ?

If $\displaystyle \frac{{df}}{{dx}}=x^x$ for x>0, and $\displaystyle g(x)=f(x^2)$ then $\displaystyle \frac{{dg}}{{dx}} =$ ?
by the chain rule: $\displaystyle \frac {dg}{dx} = f'(x^2) \cdot 2x$

now continue

3. Okay, so $\displaystyle \frac {dg}{dx} = f'(x^2) \cdot 2x$ which means that $\displaystyle \frac {dg}{dx} = x^{2x^2} \cdot 2x$
Thus $\displaystyle \frac {dg}{dx} = 2x^{2x^2+1}$
Correct?

Okay, so $\displaystyle \frac {dg}{dx} = f'(x^2) \cdot 2x$ which means that $\displaystyle \frac {dg}{dx} = x^{2x^2} \cdot 2x$
Thus $\displaystyle \frac {dg}{dx} = 2x^{2x^2+1}$