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Math Help - Derivatives of Inverse Trigonometric Functions

  1. #1
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    Derivatives of Inverse Trigonometric Functions

    Hello, and thanks in advance for your help.

    I need to find the derivative of this problem, but am at a loss as to what to do.

    h(t) = arccot(t) + arccot(1/t)

    I know that h(t) = 0 but don't quite understand how to get this answer. If anyone could show me the steps I would greatly appreciate it!
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  2. #2
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    Hello, NBrunk!

    Do you know the derivative of the inverse cotangent?
    If you do, exactly where is your difficulty?


    Differentiate: . h(t) \:=\: \text{arccot}(t) + \text{arccot}\left(\tfrac{1}{t}\right)
    . . \boxed{\text{If }y \:=\:\text{arccot}(u),\,\text{ then: }\:\frac{dy}{dx} \:=\:\frac{\text{-}\,du}{1+u^2}}


    \text{We have: }\;h(t) \;=\;\underbrace{\text{arccot}(t)}_{f(t)} + \underbrace{\text{arccot}\left(t^{-1}\right)}_{g(t)}

    . . and we want: . h'(t) \;=\;f'(t) + g'(t)



    f(t) \:=\:\text{arccot}(t)

    . . f'(t) \;=\;\frac{-1}{1+t^2}


    g(t) \;=\;\text{arccot}\left(t^{-1}\right)

    . . g'(t) \;=\;\frac{-1}{1+t^{-2}}\cdot\left(-t^{-2}\right) \;=\;\frac{1}{1 + \frac{1}{t^2}}\cdot\frac{1}{t^2} \;=\;\frac{1}{t^2+1}


    Therefore: . h'(t) \;=\;f'(t) + g'(t) \;=\;\frac{-1}{1+t^2} + \frac{1}{t^2+1} \;=\;0


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Here's an "eyeball" solution . . . which takes a bit of Thinking.


    The first term is \text{arccot}(t)

    This is an angle \theta whose cotangent is t.

    That is: . \cot\theta \:=\:\frac{t}{1} \:=\:\frac{adj}{opp}

    Its right triangle looks like this:


    Code:
                            * B
                         *  |
                      *     |
                   *        | 1
                *           |
             * θ            |
        A * - - - - - - - - *
                    t

    The second term is: . \text{arccot}\left(\frac{1}{t}\right)
    This is an angle whose cotangent is: . \frac{1}{t} \:=\:\frac{adj}{opp}

    Look at the diagram above . . . Can you find such an ngle?

    With a little mental acrobatics, we see that it is angle B, the "other" angle!


    You see: . h(t) \;=\;\text{(an angle)} + (\text{its complement}) . . . This sum is 90, a constant.


    So we did all that work to differentiate a constant!

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  3. #3
    MHF Contributor Drexel28's Avatar
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    Suppose L=\tan(x)=\frac{\sin(x)}{\cos(x)}, then \tan\left(\tfrac{\pi}{2}-x\right)=\frac{\sin\left(\tfrac{\pi}{2}-x\right)}{\cos\left(\tfrac{\pi}{2}-x\right)}=\frac{\cos(x)}{\sin(x)}=\frac{1}{\tan(x)  }=\frac{1}{L}. Therefore, \tan\left(\tfrac{\pi}{2}-x\right)=\frac{1}{L}\implies \frac{\pi}{2}-x=\arctan\left(\tfrac{1}{L}\right), but \tan(x)=L\implies x=\arctan\left(L\right) so we

    may finally conclude that \frac{\pi}{2}-\arctan\left(L\right)=\arctan\left(\tfrac{1}{L}\ri  ght). Or, more pertinently \frac{\pi}{2}=\arctan\left(L\right)+\arctan\left(\  tfrac{1}{L}\right).


    So, as Soroban pointed our we have \bigg(\arctan(x)+\arctan\left(\tfrac{1}{x}\right)\  bigg)\prime=\bigg(\tfrac{\pi}{2}\bigg)\prime=0
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