Higher order derivatives, with a twist...

Similar to the solution already posted. Consider the function:

$f(x) = x^n$ $\implies f^{1}(x) = nx^{n-1}$ $\implies f^{2}(x) = n(n-1)x^{n-2}$ $\implies f^{3}(x) = n(n-1)(n-2)x^{n-3}$

Generalising for the rth derivative:

$f^{r}(x) = n(n-1)(n-2)\dots(n-r+1)x^{n-r}$ $\implies f^{r}(x) =\frac{n!}{(n-r)!}x^{n-r}$

Using the Gamma function and the identity that:

$\Gamma\left(k+1\right)= k!$

$\implies f^{r}(x) =\frac{\Gamma\left(n+1\right)}{\Gamma\left(n-r+1\right)}x^{n-r}$

In the given case:

$n = 2 \ ; \ r = 2.5$

$\implies f^{2.5}(x) =\frac{\Gamma\left(3\right)}{\Gamma\left(\frac{1}{2}\right)}x^{-0.5}$

Knowing that $\Gamma\left(3\right)=2$ and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ , we get:

$f^{2.5}(x) = \frac{2}{\sqrt{\pi x}}$ $\implies \boxed{f^{2.5}\left(\frac{25}{\pi}\right)=\frac{2}{\sqrt{\pi \frac{25}{\pi}}}=\frac{2}{5}}$

$\large \text{Note: } \Gamma(n) = \int_{0}^{\infty} t^{n-1}e^{-t}\, dt = (n-1)!$

First of all, to find the answer, we must know that

$\large \frac{d^n}{dx^n}(x^a) = \frac{\Gamma(a+1)}{\Gamma(a-n+1)}x^{a-n}$

Using this information, we can see that $f^{(2.5)}(x) = \frac{\Gamma(2+1)}{\Gamma(2-2.5+1)}x^{2-2.5} = \frac{\Gamma(3)}{\Gamma(0.5)}x^{-0.5} = \frac{2}{\sqrt{\pi x}} \: \: \: \: \: \: \text{Note: }\Gamma(0.5) = \sqrt{\pi} \text{ and } \Gamma(3) = 2$

Now, we can plug $\frac{25}{\pi}$ in and we can see that $\frac{2}{\sqrt{\pi\left(\frac{25}{\pi}\right)}} = \frac{2}{\sqrt{25}} = \frac{\pm 2}{5} \text{ but since we want the positive value, the answer is } \boxed{\frac{2}{5} \text{ or } 0.4}$