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

1. Aug 31, 2011

### process91

1. The problem statement, all variables and given/known data
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.

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

3. 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)?

2. Aug 31, 2011

### micromass

Staff Emeritus
3. Aug 31, 2011

### process91

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.

4. Aug 31, 2011

### micromass

Staff Emeritus
Indeed!!

5. Aug 31, 2011

### process91

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

6. Aug 31, 2011

### micromass

Staff Emeritus
It's one of my favorite books as well