Question

Find the derivative of the function.
y = e^(7u) − e^(−7u)/e^(7u) + e^(−7u)

Answer 1

y = [7*e^(7u) -7*e^(-7u) - [e^(7u) * (-7)*e^(-7u) - (e^(-7u))*7*e^(7u)]]/[e^(14u)]

Answer 2

Derive the first and last terms on their own. I hope you do know that
\[\frac{ d }{ dx }e ^{x} = e^{x} ?\]
Use the chain rule as well, for the 7u.

For the middle term you will need to use the quotient rule and the chain rule.
\[\frac{ d }{ dx }[\frac{ f(x) }{ g(x) }] = \frac{ f'(x)g(x)-f(x)g'(x) }{ [g(x)]^{2} }\]

Answer 3

so \[\frac{ (7ue^{7u}+7ue ^{-7u})(e ^{7u}+e ^{-7u})-(e^{7u}-e^{-7u})(7ue^{7u}-7u^{-7u}) }{ (e^{7u} +e^{-7u})^{2}}\]
then simplify?

Answer 4

last part should be \[7ue^{-7u}\]

Answer 5

I misread your question. Didn't realize the whole thing is a quotient. Anyway, that is not quite right. What is
\[\frac{ d }{ dx } 7u ?\]

Answer 6

you know you could also simplify the whole thing to start before differentiating
\[\rightarrow \frac{e^{14u}-1}{e^{14u}+1}\]
then apply quotient rule
\[\frac{14e^{14u}(e^{14u}+1) - 14e^{14u}(e^{14u}-1)}{(e^{14u}+1)^{2}}\]
\[=\frac{28e^{14u}}{(e^{14u}+1)^{2}}\]