Question

how to find nth derivative of (e^x/a+e^-x/a)/(e^x/a-e^-x/a)

Answer 1

look for a pattern to develop

Answer 2

are those as part of exponents? or denominators?

Answer 3

ermmm exponents

Answer 4

i guess this is more clear (e^(x/a)+e^(x/a)) / (e^(x/a)-e^(-x/a))

Answer 5

the second one is -x/a made a mistake dere

Answer 6

the only saving grace might be that we are playing with e's

Answer 7

\[f = (e^{x/a}+e^{x/a})~ (e^{x/a}-e^{x/a})^{-1}\]
\[f' = \frac1a(e^{x/a}+e^{x/a})~ (e^{x/a}-e^{x/a})^{-1}-\frac1a(e^{x/a}-e^{x/a})(e^{x/a}+e^{x/a})~ (e^{x/a}-e^{x/a})^{-2}\]

\[f' = \frac{\frac1a(e^{x/a}+e^{x/a})~ (e^{x/a}-e^{x/a})-\frac1a(e^{x/a}-e^{x/a})(e^{x/a}+e^{x/a})}{(e^{x/a}-e^{x/a})^{2}} \]

\[f' = \frac{\frac1a(e^{x/a}-e^{x/a})~ (e^{x/a}-e^{x/a})-\frac1a(e^{x/a}-e^{x/a})(e^{x/a}+e^{x/a})}{(e^{x/a}-e^{x/a})^{2}} \]

does that zero out?

Answer 8

it might be simpler to run thru these using the wolf

change "first" to second, third, 4th etc to see if you can develop a pattern

http://www.wolframalpha.com/input/?i=first+derivative++%28e%5E%28x%2Fa%29%2Be%5E%28-x%2Fa%29%29+%2F+%28e%5E%28x%2Fa%29-e%5E%28-x%2Fa%29%29

Answer 9

hmm there are formulas to find out nth derivative of each exponential functions

Answer 10

i just wanted to know if u can expand it in one simple line and then solve it

Answer 11

simple being a relative term i spose ;)