Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One popular proof is to take $\sin{y} = x$ and then differentiate on both sides. But how do you prove it from first principles? Help very much appreciated.

share|cite|improve this question
Why isn't that proof from first principles? It's a general principle that if you know the derivative of a function, and it has an inverse, then you know the derivative of the inverse, just by the chain rule. – Ben Millwood Jun 4 '12 at 15:16

Is this cheating? We want the limit as $h$ approaches $0$ of $\frac{\arcsin h-0}{h}$. Let $w=\arcsin h$. So we are interested in the limit of $\frac{w}{\sin w}$ as $w$ approaches $0$. Upside down, but familiar!

Now we know the derivative at $0$. We can get the derivative at $x$ by using the $\arcsin$ version of the addition law for sines.

share|cite|improve this answer
+1) No that is not cheating, sine must come into play somehow. – AD. Jun 4 '12 at 15:14
@AD.: One can imagine defining $\arcsin$ first, for example as an integral. Just like with exponential/logarithm, we can take one or the other as fundamental. – André Nicolas Jun 4 '12 at 15:18
Ok, I see your point. Personally, I like to define $\arcsin$ as the inverse of $\sin$ around 0. – AD. Jun 4 '12 at 15:21
It's not cheating, but it only gives the derivative at $x=0$... – Hans Lundmark Jun 4 '12 at 20:50

We may define $\displaystyle\sin^{-1} x$ as $\displaystyle\int^x_0 \frac{1}{\sqrt{1-t^2}}dt$. Then by differentiating both sides, we get that $\displaystyle(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}}.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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