simplification of square root of $\pi$

Question

Using the power rule, my textbook differentiates this:

$\frac{d}{dx}(\sqrt{x^{2+\pi}})$

like this, using the power rule:

$$\begin{align} =& \frac{d}{dx}(x^{1+(\pi/2)}) \tag{1}\\ =& (1+\frac{\pi}2)x^{1+(\pi/2)-1} \tag{2}\\ =& \frac12(2+\pi)\sqrt{x^{\pi}} \tag{3} \end{align}$$

I understand how we went from (1) to (2), but I don't understand how $x^{2+\pi}$ became $(x^{1+(\pi/2)}$ in 1, since the square root of $\pi \neq \pi/2$.

I'm also lost with regards to the steps between $2$ and $3$.

Answer 1

$$\sqrt{x^{2+\pi}}=x^{\frac{2+\pi}2}=x^{1+\frac{\pi}{2}}$$

Answer 2

No its $x^{(2+\pi).0.5}=x^{1+(\pi/2)}$