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Math Help - What is the derivative of arccot(e^2x)?

  1. #1
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    What is the derivative of arccot(e^2x)?

    What is the derivative of arccot(e^2x)? I did this problem by switching the expression to arctan[e^(-2x)] then got (-2e^-2x)/(e^-4x + 1) but my book says (-2e^2x)/(1 + e^4x).

    What am I doing wrong?
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  2. #2
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    Let y=\cot^{-1}(e^{2x}), then \cot(y)=e^{2x}, \tan(y)=e^{-2x}

    Now differentiate both sides wrt x:

    \sec^2(y)\frac{dy}{dx}=-2e^{-2x} and so \frac{dy}{dx}=\frac{-2e^{-2x}}{\sec^2(y)}

    Using \sec^2(y)=1+\tan^2(y) = 1+e^{-4x} we have

    \frac{dy}{dx}=\frac{-2e^{-2x}}{1+e^{-4x}}

    Now multiply top and bottom by e^{4x} to get \frac{-2e^{2x}}{e^{4x}+1}.
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  3. #3
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    Also, y=\text{arccot}\left(e^{2x}\right)=\arctan\left(e^  {-2x}\right) \Rightarrow y'=\frac{-2e^{-2x}}{e^{-4x}+1}=\frac{-2e^{2x}}{e^{4x}+1}.
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  4. #4
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    Hey Abu-Khalil! Thanks a lot, I don't know why I didn't recognize the fact that I could've simplified the expression by multiplying the numerator and the denominator by e^4x. At least I technically answered correctly.
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