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, 08:07 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, 09:00 AMThomas154321
Let , then

Now differentiate both sides wrt x:

and so

Using we have

Now multiply top and bottom by e^{4x} to get . - Feb 25th 2009, 09:37 AMAbu-Khalil
Also, .

- Feb 25th 2009, 09:54 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)