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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- Feb 25th 2009, 07:07 AMKaitosanWhat 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? - Feb 25th 2009, 08:00 AMThomas154321
Let $\displaystyle y=\cot^{-1}(e^{2x})$, then $\displaystyle \cot(y)=e^{2x}, \tan(y)=e^{-2x}$

Now differentiate both sides wrt x:

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

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

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

Now multiply top and bottom by e^{4x} to get $\displaystyle \frac{-2e^{2x}}{e^{4x}+1}$. - Feb 25th 2009, 08:37 AMAbu-Khalil
Also, $\displaystyle y=\text{arccot}\left(e^{2x}\right)=\arctan\left(e^ {-2x}\right)$ $\displaystyle \Rightarrow y'=\frac{-2e^{-2x}}{e^{-4x}+1}=\frac{-2e^{2x}}{e^{4x}+1}$.

- Feb 25th 2009, 08:54 AMKaitosan
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. (Talking)