# 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

## 2 Answers

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 $$\frac{\pi}{2}$$.

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.