0
$\begingroup$

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

$\endgroup$
3
3
$\begingroup$

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

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.