# Derivative of arccos (+ feedback :D)

• June 10th 2011, 03:06 AM
TeaWithoutMussolini
Derivative of arccos (+ feedback :D)
So I posted a lot of questions last night and this morning, I want to say thanks to all those who helped me. I am just back from the exam and yes most of the advice I got from you all was relevant to what came up in it... I believe I will have passed well ONLY because of the advice I got on here. One night of topic questions on here and I learn WAY more than one whole year of university tuition... well that's another rant for another day :) But anyway the point I want to make and the question I want to ask is:

Last night I asked for help on $y=arctan(x)$ and finding $\frac{dy}{dx}$. In this exam, the similar thing came up this time $y=arccos(x)$. It is the only question I did not know and I should have been prepared for this... but, I could not find a substitution like the one used for the arctan. Now please somebody tell me what I can do once I get to $\frac{-1}{sin(y)}$, remember that $x=cos(y)$ this time and I had to find something, I think, to link sin and cos, but I couldn't do it? I know you can't just square them otherwise its easy...
• June 10th 2011, 03:51 AM
HallsofIvy
So you got $\frac{-1}{sin(arccos(x))}$.

There are two ways to think about this- the more formal is to use trig identities. You know, of course, that $sin^2(t)+ cos^2(t)= 1$ so that $sin(t)= \pm\sqrt{1- cos^2(t)$. Replacing t by arccos(x), $sin(arccos(x))= \pm\sqrt{1- cos(arccos(x))^2}$ and, of course, $cos(arccos(x))= x$. That is, $\frac{-1}{sin(arccos(x))}= \fra{\pm 1}{\sqrt{1- x^2}}$. Which sign is appropriate depends upon which quadrant you are in.

Less formal but perhaps simpler to see- imagine a right triangle with one angle y, "near side" of length x, and hypotenuse of length 1 (so that cos(y)= x/1= x). By the Pythagorean theorem, the "opposite side" has length $\sqrt{1- x^2}$ so that $sin(y)= \sqrt{1- x^2}$ and them $\frac{-1}{sin(y)}= \frac{-1}{\sqrt{1- x^2}}$ which doesn't have the sign ambiguity because in setting up that right triangle we are assuming y is in the first quadrant.