# Inverse Trig Function Derivative (Apostol Section 6.22 #11)

## Homework Statement

Given that $\frac{d}{dx} (\text{arccot}{x}-\arctan{1/x})=0 \hspace{10mm} \forall x \ne 0$,
prove that there is no constant C such that $\text{arccot}{x}-\arctan{\frac{1}{x}}=C \hspace{10mm} \forall x \ne 0$
and explain why this does not contradict the zero-derivative theorem.

## Homework Equations

The Zero-Derivative Theorem:
If f'(x) = 0 for each x in an open interval I, then f is constant on I.

## The Attempt at a Solution

The first part of this problem has you verify that the derivative is indeed zero, which I did verify. I think that $\text{arccot}{x}-\arctan{\frac{1}{x}}=0 \hspace{10mm} \forall x \ne 0$, however, since $\text{arccot}{x}=y \implies x=\cot{y} \implies \frac{1}{x} = \tan{y} \implies \arctan{\frac{1}{x}}=y$.

WolframAlpha seems to agree:
http://www.wolframalpha.com/input/?i=arccot(x)+-+arctan(1/x)

So is Apostol not considering 0 a constant (that is, when he refers to "a constant C", is C necessarily not equal to 0)?

OK, so the reason my "proof" that it is equal to zero doesn't work is in the last step, where $\arctan{\frac{1}{x}}$ may equal y (if $y\in(0,\frac{\pi}{2})$), or it may equal $y-\pi$, correct?

Then the reason that this is not a contradiction of the zero derivative theorem is that at x=0 there is a discontinuity, and on either side of 0 it would apply but the constants are not equal.

Indeed!!

Thanks for your help! I'm loving Apostol's book so far.

Thanks for your help! I'm loving Apostol's book so far.

It's one of my favorite books as well