Find a 'formula' (bad translation?) of the $n$-timed ($n\in \mathbb N$) derivative of $\sqrt x$, and prove it's correct for all $n$ with induction. What I found: $$({1\over2})^{n-1}|{3\over2}-n|(-1)^{n+1}\over{x^{n-1/2}}$$
Is it correct? If so, how do I show it is by induction?
Let $f(x) = x^{1/2}$, then
So maybe $$\large f^{(n)}(x) = (-1)^?\frac{?}{2^n}x^{\frac{-2n+1}{2}}. $$ You can probably find the missing pieces.
As for proving that this formula is true, you could use induction: As shown above the formula is true for $n=1$. So assume that it is true for $n$ and prove it for $n+1$. So you get $$\large f^{(n+1)}(x) = \frac{d}{dx}[(-1)^?\frac{?}{2^n}x^{\frac{-2n+1}{2}}] = \dots = (-1)^?\frac{?}{2^{n+1}}x^{\frac{-2(n+1) + 1}{2}}. $$ Again, you can probably fill in the missing details. And once you have gotten this far you are done.
Each time you derive you get the power down, and the powers are:
$$\frac{1}{2}..\frac{-1}{2}..\frac{-3}{2}...$$
The denominator is $2^k$, and the numerator is $\prod_{i=1}^k(2i-3)$, plus a $(-1)^{k+1}$ for the $-$ signs. finally, the powers of $x$ are reduced in $1/2$ every time. the final formula is:
$$(\sqrt{x})^{(k)}=\frac{(-1)^{k+1}(\prod(2i-3))x^{\frac{1-2k}{2}}}{2^k}$$
