Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse Trig Function Derivative (Apostol Section 6.22 #11)

  1. Aug 31, 2011 #1
    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:

    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. jcsd
  3. Aug 31, 2011 #2
  4. Aug 31, 2011 #3
    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.
  5. Aug 31, 2011 #4
  6. Aug 31, 2011 #5
    Thanks for your help! I'm loving Apostol's book so far.
  7. Aug 31, 2011 #6
    It's one of my favorite books as well :biggrin:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook