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$f(x)=\arcsin(x)$

$f'(x)=\frac{1}{f^{-1}(x)}$

$f'(\arcsin(x))=\frac{1}{\sin'x}=\cos^{-1}x$

$(f^{(n-1)})'(x)=(\cos^{-1}x)^{(n-1)}$

It's a complex function: $x$ itself, $\cos x$, and $^{-1}$. Let's use the chain rule.

$(\cos^{-1}x)^{(n-1)}=\cos (x+\frac{(n-1)\pi}{2})\cdot (-1 \cdot -2 \cdot \cdots \cdot (-1-n+1+1))\cos^{-1-n+1}x=$

$=\cos (x+\frac{(n-1)\pi}{2})\cdot (-1)^{n-1}\cdot (n-1)!\cdot \cos^{-n} x$

$f^{(n)}(0)=\cos (\frac{(n-1)\pi}{2})\cdot (-1)^{n-1}\cdot (n-1)!$

If $n$ is even, then the cosine turns into zero, and the whole derivative becomes zero.

If $n$ is odd, then

$f^{(2k+1)}(0)=\cos (\frac{(2k)\pi}{2})\cdot (2k)!=(-1)^{k}\cdot (2k)!$.

The book, however, gives $(1\cdot 3\cdot \cdots \cdot (2k-1))^2$ for the odd case.

Do I misunderstand something fundamentally?

Noa Even
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fragileradius
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2 Answers2

3

You’re using the incorrect formula for the derivative of an inverse function.

You wanted to use

$$(f^{-1}(x))’ = \frac{1}{f’(f^{-1}(x))}$$

Without using the formula, all you really need is the Chain Rule.

$$y = \arcsin x \iff \sin y = x$$

$$\frac{d}{dx} \sin y = \frac{d}{dx}x$$

$$\frac{du}{dx} = \frac{du}{dy}\cdot\frac{dy}{dx} \implies \cos y\cdot\frac{dy}{dx} = 1$$

$$\frac{dy}{dx} = \frac{1}{\cos y} \implies \frac{dy}{dx} = \frac{1}{\color{blue}{+}\sqrt{1-\sin^2y}}$$

$$\frac{dy}{dx} = \frac{1}{\color{blue}{+}\sqrt{1-\sin^2(\arcsin x)}} = \frac{1}{\color{blue}{+}\sqrt{1-x^2}}$$

Note the positive sign as $\cos y$ is positive in the range of $\arcsin x$, which is $\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]$.

KM101
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1

Your way to proceed is really not clear, the standard way is as follows

$$y=\arcsin x \implies \sin y = x \implies \cos y\cdot y'=1 \implies y'=\frac1{\cos y}$$

and since $\cos y$ is positive in the range of $\arcsin x$

$$\cos y= \cos (\arcsin x)=\sqrt{1-\sin^2(\arcsin x)}=\sqrt{1-x^2}$$

user
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