differentiate (e^2u)/(e^u+e^-u)?

Answer 1

D(e^2u) = 2e^2u D(e^u + e^-u) = e^u - e^-u

D((e^2u)/(e^u + e^-u))
= (2e^2u(e^u + e^-u) - e^2u(e^u - e^-u))/(e^u + e^-u)^2
=(2e^3u + 2e^u - e^3u + e^u)/(e^2u + e^-2u +2)
=(e^3u + 3e^u)/(e^2u + e^-2u + 2)

Answer 2

Here's logarithmic differentiation. We haven't learned it in my calculus class yet. I taught myself, so it's probably wrong:

y = (e^(2u)) / (e^u+e^(-u))

ln y = ln ((e^(2u)) / (e^u + e^(-u)))

ln y = ln e^u + ln e^u - ln (e^u + e^(-u))

ln y = u*ln e + u*ln e - ln (e^u + e^(-u))

ln y = 2u - ln (e^u + e^(-u))

1/y * dy/dx = 2 - (1/e^u + e^(-u))(e^u - e^(-u))

1/y * dy/dx = 2 - (e^u - e^(-u))/(e^u + e^(-u))

dy/dx = (2 - (e^u - e^(-u))/(e^u + e^(-u)))((e^(2u)) / (e^u + e^(-u)))

dy/dx = (2e^(2u) / (e^u + e(-u))) - ((e^(3u) - e^u) / ((e^u + e^(-u))^2))

Answer 3

you need the division rule and logarithmic differentiation to do this one:
(der. of numerator)(denominator) - (der. denominator)(num)
-----------------------------------------------------------------------------
(denominator)^2

der of numerator = 2e^2u
der of denominator = e^u - e^u = 0

figure the rest out hah.

Answer 4

something I don't understand / something I will never understand