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.
$f^{(k+1)}(x) = f^{(k)'}(x)$
 A: You start with letting $A=\frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)$ (a constant), so
\begin{align*}
f^{(k)}(x) & = A\cdot\frac{1}{\sqrt{x^{2k-1}}}\\
f^{(k+1)}(x) & = A\cdot\frac{d}{dx}\frac{1}{\sqrt{x^{2k-1}}}\\
& = A\cdot\frac{d}{dx}x^{\frac{1-2k}{2}}\\
& = A\cdot \left(\frac{1-2k}{2}\right) \cdot x^{\frac{1-2k-2}{2}}\\
& = \color{red}{A}\cdot \left(\frac{1-2k}{2}\right) \cdot \frac{1}{\sqrt{x^{2(k+1)-1}}}\\
&= \color{red}{\frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)} \left(\frac{1-2k}{2}\right) \cdot \frac{1}{\sqrt{x^{2(k+1)-1}}}\\
&= \frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1) \color{red}{(-1)\left(\frac{2k-1}{2}\right)} \cdot \frac{1}{\sqrt{x^{2(k+1)-1}}}\\
&= \color{blue}{\frac{(-1)^{k+2}}{2^{k+1}}\cdot\prod^{k}_{j=1}(2j-1)} \cdot \frac{1}{\sqrt{x^{2(k+1)-1}}}.
\end{align*}
A: If$$f^{(k)}(x)=\frac{(-1)^k}{2^k}\prod_{j=1}^{k-1}(2j-1)\frac1{\sqrt{x^{2k-1}}}=\frac{(-1)^k}{2^k}\prod_{j=1}^{k-1}(2j-1)x^{1/2-k},$$then\begin{align}f^{(k+1)}(x)&=\frac{(-1)^k}{2^k}\left(\prod_{j=1}^{k-1}(2j-1)\right)\left(\frac12-k\right)x^{1/2-k-1}\\&=-\frac12\frac{(-1)^k}{2^k}\left(\prod_{j=1}^{k-1}(2j-1)\right)\left(2k-1\right)x^{1/2-(k+1)}\\&=\frac{(-1)^{k+1}}{2^{k+1}}\prod_{j=1}^k(2j-1)\frac1{\sqrt{x^{2(k+1)-1}}}.\end{align}
A: 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
$$f^{(k+1)}(x)=\frac{(-1)^k}{2^{k+1}}\prod^{k-1}_{j=1}(2j-1)\cdot(2k-1)\ \frac1{x^{\frac{2k-1}2+1}}.$$
