<s>[INST] Prove derivative by induction $f:(0, \infty) \rightarrow \mathbb{R}$
$f(x) = \sqrt{x}$
a) Calculate the first four derivatives
$f'(x) = \frac{1}{2}\cdot \frac{1}{\sqrt{x}}$
$f''(x) = -\frac{1}{4}\cdot \frac{1}{\sqrt{x^3}}$
$f'''(x) = \frac{3}{8}\cdot \frac{1}{\sqrt{x^5}}$
$f''''(x) = -\frac{15}{16}\cdot \frac{1}{\sqrt{x^7}}$
b) Prove by induction that the following formula holds true:
$f^{(k)}(x) = \frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}$
Base Case: $k=1$:
$f'(x) = \frac{(1)}{2}\cdot\prod^{0}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x}} = \frac{1}{2}\cdot\frac{1}{\sqrt{x}}$
Inductive Hypothesis(IH):
Assumption holds true for some k.
Inductive step:
$k \rightarrow k+1$
to show:
$f^{(k+1)}(x) = \frac{(-1)^{k+2}}{2^k+1}\cdot\prod^{k}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k+1}}}$
$\frac{(-1)^{k+2}}{2^k+1}\cdot\prod^{k}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k+1}}} \\
= \frac{(-1)^{k+1}}{2^k}\cdot\frac{-1}{2} \cdot(2k-1)\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}\cdot \frac{1}{\sqrt{x^2}} \\
= [\frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}] \cdot\frac{-1}{2}\cdot(2k-1)\frac{1}{\sqrt{x^2}} \\
=^{IH} f^{(k)}(x) \cdot \frac{-1}{2}\cdot(2k-1)\frac{1}{\sqrt{x^2}} \\
= f^{(k)}(x) \cdot (-k+\frac{1}{2})\cdot \frac{1}{|x|} \text{since x $\in$ $(0, \infty)$} \\
= f^{(k)}(x) \cdot [(-k+\frac{1}{2})\cdot \frac{1}{x}]$
This means that to get from one derivative to the next, the factor at the end will be multiplied.
So how do I go on from here?
Usually I would start from the left side. But if I would start here from the left side I would have to transform $f^{(k+1)}(x)$. And I would have to calculate the derivative over the product symbol.
$f^{(k+1)}(x) = f^{(k)'}(x)$
 [/INST] Hint:
First, I would write the formula in a slightly different way:
$$f^{(k)}(x) = \frac{(-1)^{k\color{red}-1}}{2^k}\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}$$
Next, to differentiate $f^{(k)}(x)$, I would use a fractional writing of the exponent of $x$, and use the formula
$$\Bigl(\frac1{x^\alpha}\Bigr)'=-\frac \alpha{x^{\alpha+1}}\quad\text{ with }\quad \alpha=\tfrac{2k-1}2.$$
This yields readily
$$f^{(k+1)}(x)=\frac{(-1)^k}{2^{k+1}}\prod^{k-1}_{j=1}(2j-1)\cdot(2k-1)\ \frac1{x^{\frac{2k-1}2+1}}.$$
</s>