Inverse Trig Function Derivative (Apostol Section 6.22 #11)

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Homework Statement


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.


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 [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)?
 

Answers and Replies

  • #3
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OK, so the reason my "proof" that it is equal to zero doesn't work is in the last step, where [itex]\arctan{\frac{1}{x}}[/itex] may equal y (if [itex]y\in(0,\frac{\pi}{2})[/itex]), or it may equal [itex]y-\pi[/itex], 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
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Indeed!!
 
  • #5
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Thanks for your help! I'm loving Apostol's book so far.
 
  • #6
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Thanks for your help! I'm loving Apostol's book so far.

It's one of my favorite books as well :biggrin:
 

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