Inverse Trig Function Derivative (Apostol Section 6.22 #11)

[process91]

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)?

 

[micromass]

I think the issue here is that wolfram alpha and Apostol defined the acot(x) in different ways.

Wolfram alpha has this graph as acot(x): http://www.wolframalpha.com/input/?i=y=acot(x)
While Apostol defines acot(x) such that this graph is obtained: http://www.wolframalpha.com/input/?i=y=pi/2-atan(x)

So the graph of acot(x)-atan(1/x) how Apostol wants it is http://www.wolframalpha.com/input/?i=y=pi/2-atan(x)-atan(1/x)

 

[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.

 

[micromass]

Indeed!

 

[process91]

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

 

[micromass]

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

It's one of my favorite books as well