(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given that [itex]\frac{d}{dx} (\text{arccot}{x}-\arctan{1/x})=0 \hspace{10mm} \forall x \ne 0[/itex],

prove that there is no constant C such that [itex]\text{arccot}{x}-\arctan{\frac{1}{x}}=C \hspace{10mm} \forall x \ne 0[/itex]

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 [itex]\text{arccot}{x}-\arctan{\frac{1}{x}}=0 \hspace{10mm} \forall x \ne 0[/itex], however, since [itex]\text{arccot}{x}=y \implies x=\cot{y} \implies \frac{1}{x} = \tan{y} \implies \arctan{\frac{1}{x}}=y[/itex].

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)?

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

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