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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?

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2 Answers 2

Let $f(x) = x^{1/2}$, then

  • $f'(x) = \frac{1}{2}x^{-1/2}$
  • $f''(x) =\frac{1}{2}\frac{-1}{2}x^{-3/2}$
  • $f^{(3)}(x) = \frac{1}{2}\frac{-1}{2}\frac{-3}{2}x^{-5/2}$
  • $f^{(4)}(x) = \frac{1}{2}\frac{-1}{2}\frac{-3}{2}\frac{-5}{2}x^{-7/2}$

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.

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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:

$$(\sqrt{x})^{(k)}=\frac{(-1)^{k+1}(\prod(2i-3))x^{\frac{1-2k}{2}}}{2^k}$$

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