Why is the derivative of $\arctan(\frac{x}{\sqrt{1-x^2}})$ the same as the derivative of $\arcsin(x)$? I'm self studying math through MIT opencourseware. Unfortunately the solutions of the problem sets are not very user-friendly and don't contain much explanation. There is a question which is bugging me and I couldn't come up with any explanation (there is no explanation provided in the solutions as well).
The question is:
Why is the derivative of $\arctan(\frac{x}{\sqrt{1-x^2}})$ the same as the derivative of $\arcsin(x)$?
I have solved the derivative of the arctan part and it's obvious to me how to get to the $\frac{1}{\sqrt{1-x^2}}$ answer. But I don't understand how this is related to the $\arcsin(x)$. Why are they the same?
Thank you in advance
 A: For $\alpha$ in the first quadrant, if $\sin(\alpha)=x$, then $\cos(\alpha)=\sqrt{1-x^2}$, and so
$$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{x}{\sqrt{1-x^2}}.$$
That means that if $\arcsin(x)=\alpha$, then $\arctan\left(\frac{x}{\sqrt{1-x^2}}\right) = \alpha$ as well. That is, $\arcsin(x) = \arctan\left(\frac{x}{1-x^2}\right)$. They are the same function (at least on $[0,1)$), so they have the same derivative.
A similar result holds for $\alpha$ in the fourth quadrant, which are the angles you get when $x$ is negative: you still have $\cos(\alpha)=\sqrt{1-\sin^2(\alpha)}$, because the cosine is nonnegative for angles in $[-\frac{\pi}{2},\frac{\pi}{2}]$.
Since they are in fact the same function where they are both defined (which is $-1\lt x\lt 1$), it is not a surprise they have the same derivative.
More generally, two functions have the same derivative exactly when (if and only if) they differ by a constant (this is the Constant Function Theorem). So for example, $-\arctan(x)$ and $\arctan(\frac{1}{x})$ (for $x\gt 0$) have the same derivative:
$$\begin{align*}
-\frac{d}{dx}\arctan(x) &= -\frac{1}{1+x^2}\\
\frac{d}{dx}\arctan\left(\frac{1}{x}\right) &= \left(\frac{1}{1+\frac{1}{x^2}}\right)\left(\frac{1}{x}\right)’\\
&= \frac{1}{\quad\frac{x^2+1}{x^2}\quad}\left(-\frac{1}{x^2}\right)\\
&= -\frac{1}{x^2+1}.
\end{align*}$$
so $-\arctan(x)$ and $\arctan\left(\frac{1}{x}\right)$ differ by a constant on $(0,\infty)$. You can figure out the value since at $x=1$ you get $-\arctan(1) = -\frac{\pi}{4}$, and $\arctan(\frac{1}{1}) = \frac{\pi}{4}$, so the two functions differ by $\frac{\pi}{2}$.
A: Conceptually, both $\theta_t(x)=\arctan\frac{x}{\sqrt{1-x^2}}$ and  $\theta_s(x)=\arcsin(x)$ represent angles as functions of $x$. Equivalently, the two expressions are $\tan( \theta_t (x))= \frac{x}{\sqrt{1-x^2}}$ and $\sin(\theta_s(x))=x$, respectively. Then, $\cos (\theta_s(x))= \sqrt{1-x^2}$ and
$$\tan(\theta_s(x) )= \frac {\sin(\theta_s(x))}{\cos(\theta_s(x))}= \frac{x}{\sqrt{1-x^2}}=\tan( \theta_t(x))
$$
Thus, $\theta_t(x)$ and $\theta_s(x)$ are the same. So are their derivatives.
