Asked by TayB
Find the derivative of the function.
y=(e^3u -e^-3u)/(e^3u+e^-3u)
y=(e^3u -e^-3u)/(e^3u+e^-3u)
Answers
Answered by
Steve
the easy way is to notice that this is just
y = tanh(3u)
y' = -3sech^2(3u)
or, using the exponentials, that's
12/(e^3u + e^-3u)^2
or
(12e^6u)/(e^6u + 1)^2
y = tanh(3u)
y' = -3sech^2(3u)
or, using the exponentials, that's
12/(e^3u + e^-3u)^2
or
(12e^6u)/(e^6u + 1)^2
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