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Math Help - Calculus Theory:differentiating

  1. #1
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    Calculus Theory:differentiating


    Verify the identity:

    arctanx+ arctan (1/x)=( ∏/2)

    using calculus theory.

    differentiate the left hand side of the identity.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Calculus Theory:differentiating

    If f(x)=\arctan x+\arctan (1/x)\; (x\neq 0) then, f'(x)=\ldots=0 . As (0,+\infty) is connected, then f(x) is constant in (0,+\infty). But f(1)=\pi/4+\pi/4=\pi/2 so f(x)=\pi/2 in (0,+\infty) . Similar arguments for (-\infty,0) .
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    Re: Calculus Theory:differentiating

    I don't get it.
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    Re: Calculus Theory:differentiating

    Quote Originally Posted by LAPOSH42 View Post
    I don't get it.
    The derivative f'(x)=\frac{1}{1+x^2}+\frac{-1}{x^2+1}=0 for all x\ne 0.
    So the function must be constant on its domain.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Calculus Theory:differentiating

    Quote Originally Posted by Plato View Post
    The derivative f'(x)=\frac{1}{1+x^2}+\frac{-1}{x^2+1}=0 for all x\ne 0.
    So the function must be constant on its domain.
    Take as a counterexample:

    f(x)=\begin{Bmatrix} 1 & \mbox{ if }& x<0\\0 & \mbox{if}& x>0\end{matrix}

    Then, f'(x)=0 for all x on its domain and f(x) is not constant. As I said in answer #2 the domain must be connected, or equivalently (in \mathbb{R}) an interval.
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    Re: Calculus Theory:differentiating

    Quote Originally Posted by FernandoRevilla View Post
    Take as a counterexample:

    f(x)=\begin{Bmatrix} 1 & \mbox{ if }& x<0\\0 & \mbox{if}& x>0\end{matrix}
    Then, f'(x)=0 for all x on its domain and f(x) is not constant. As I said in answer #2 the domain must be connected, or equivalently (in \mathbb{R}) an interval.
    Yes, I understand that. It was a quick reply to that particular problem. I should have mentioned the two different connected domain (0,\infty).

    If fact, the function is \frac{-\pi}{2} on (-\infty,0)
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