Differentiate $\arcsin(\frac{1}{\sqrt{1 + x^2}})$ This is from Calculus by Michael Spivak, 3rd Edition, Chapter 15, problem 1 (iv):

Differentiate $f(x) = \arcsin\left(\frac{1}{\sqrt{1 + x^2}}\right).$

Here's my attempt:
\begin{align}f'(x) &= \left[\arcsin\left(\frac{1}{\sqrt{1 + x^2}}\right)\right]' \\
&= \arcsin'\left(\frac{1}{\sqrt{1 + x^2}}\right)\cdot \left(\frac{1}{\sqrt{1 + x^2}}\right)' \\
&= \frac{1}{\sqrt{1 - \left(\frac{1}{\sqrt{1 + x^2}}\right)^2}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{1}{\sqrt{1 - \frac{1}{1+x^2}}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{1}{\sqrt{\frac{1 + x^2 - 1}{1+x^2}}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{-x}{\frac{\sqrt{x^2}}{\sqrt{1+x^2}} \cdot \sqrt{1 + x^2}(1+x^2)} \\
&=\frac{-x}{|x|} \cdot \frac{1}{1+x^2}.\end{align}
Therefore
$$f'(x) =
\begin{cases}
\dfrac{-1}{1+x^2},  & \text{if $x > 0$} \\[1ex]
\dfrac{1}{1+x^2}, & \text{if $x < 0$}
\end{cases}
$$
In the Answer Book the published solution is $\frac{-1}{1+x^2}$.
It seems Spivak is discounting the possibility that $x < 0$. Why might this be? Is there a reason $x$ cannot be negative here? Did I make a mistake? Perhaps with this part of simplifying the denominator?
$$\sqrt{\frac{x^2}{1+x^2}} \cdot \sqrt{1 + x^2} = \sqrt {x^2} = |x|$$
I suspect there's something simple I'm not seeing. I'm just now encountering these inverse trigonometric functions, so my understanding isn't very solid yet.
Edit: This error could have become a misleading, possibly demoralizing waste of time, but thanks your insightful answers it's instead proven to be more instructive than the first 3 (omitted, correct) parts of the problem.
 A: Your argument is fine, and it appears that the Answer Book (assuming that you are quoting it correctly—I don't have access to that book) is incorrect (or, at least, omitting a hypothesis or restriction of some sort).  That being said, the real issue here is not that the solutions manual is wrong, but that you lack confidence in your work, or lack the resources to reassure yourself that what you have done is correct.
This is fine!
Lacking confidence is a part of learning.  However, another part of learning is figuring out how to acquire confidence in your own work, and what kind of resources might help.  So, rather than commenting on the particular problem that you are facing, allow me to address the larger question:  if I believe that at a text is wrong, what can I do?

*

*First and foremost, reread the text.  Then reread it again.  Make sure that you are not missing any important definitions or ideas.  In this case, I would wonder about the domain of the function being considered, and how Spivak treats the derivative of a function which is not differentiable on its domain.
In this case, the function you are differentiating is not differentiable at zero.  It might be reasonable to conclude that Spivak is only giving the derivative on some domain where the function is differentiable, e.g. on $(0,\infty)$.  You might look back at earlier chapters and see if this might provide an explanation.
Of course, if this is the case, I would argue that Spivak is being a little sloppy, but pobody's nerfect.


*Look for a list of errata (an erratum is an error made up to the time of printing, such as a typographical error).  When mistakes or errors are made in texts—particularly technical texts such as mathematics or physics books—the author or publisher (or an interested third party) may keep and publish a list of errata.  If you think that there is an error in the text, try looking for such a list.
For example, the Google search spivak calculus third edition errata may bring up some useful results.  None of the first page of results looks overly promising, but your milage may vary.


*Look for alternative ways of understanding the problem and/or verifying that your solution is correct.  If you are trying to find an antiderivative, try differentiating to check your work.  Use a CAS (computer algebra system) to check your solution.  Look for patterns or symmetries which might help you determine whether you are right or wrong.
In this case, the function
$$ x \mapsto \frac{1}{\sqrt{1+x^2}} $$
is even.  It is not terribly difficult to verify that the composition of any function with an even function is again even, and so
$$x \mapsto \arcsin\!\left( \frac{1}{\sqrt{1+x^2}} \right) $$
must also be even.  A useful fact about even functions is that their derivatives are odd, hence the expectation is that the result here should be an odd function.  Spivak's solution is not odd, while yours is.  This should give you some confidence (at the very least, you know that Spivak has been unclear or made an error).
Another possibility is to use some technology to build confidence.  For example, GeoGebra can be used to graph Spivak's function, compute derivatives, and so on.  A graph of GeoGebra's derivative and your own should be convincing.  Again, this is not proof that you are right, but should help you to have confidence in your result.
A: Observe that
$$f(\sinh(x))=\arcsin(\frac{1}{\sqrt{1+\sinh^2(x)}})=\arcsin(\frac{1}{\cosh(x)})$$
The differentiation gives
$$f'(\sinh(x))\cosh(x)=$$
$$\frac{1}{\sqrt{1-\frac{1}{\cosh^2(x)}}}\frac{-\sinh(x)}{\cosh^2(x)}=$$
$$\frac{-1}{\cosh(x)}\frac{\sinh(x)}{|\sinh(x)|}$$
$$\implies f'(\sinh(x))=\frac{\pm 1}{\cosh^2(x)}=\frac{-sign(x)}{1+\sinh^2(x)}$$
So,
$$f'(X)=\frac{-sign(X)}{1+X^2}$$
A: Let us verify in another way:
Let $y=\arcsin\dfrac1{\sqrt{1+x^2}}\implies\dfrac\pi2\ge y\ge0$
and $x^2=\csc^2y-1$
If $x\ge0, y=\text{arccot}x=\dfrac\pi2-\arctan x$
If $x<0, x=-\cot y, $
Consequently, $y=\cdots=\dfrac\pi2-\arctan(-x)=\dfrac\pi2+\arctan x$
