Find dy/dx how would i do this? y=a/2(e^x/a - e^-x/a)?

Answer 1

The answer posted is correct, but it seems rather silly that they gave you this question where you had to simplify the 'a' out first before finding the derivative, so I am going to assume that the question should be stated as follows:
Find dy/dx. y=a/2(e^[x/a] - e^[-x/a])? - this makes more sense to ask this question and I think this is what they wanted you to find.

Since d[e^u]/dx = [e^u]*[du/dx]
Hence, where u = x/a, and du/dx = (1/a)
d[e^[x/a]]/dx = (1/a)*e^[x/a]

and now let u = -x/a and du/dx = -(1/a)
d[e^[-x/a]]/dx = (-1/a)*e^[-x/a]

Therefore, dy/dx = [a/2]*[d[e^[x/a]]/dx - d[e^[-x/a]]/dx]
dy/dx = [a/2]*[(1/a)*e^[x/a] - (-1/a)*e^[-x/a]]
= [a/2]*[(1/a)*e^[x/a] + (1/a)*e^[-x/a]]
= [a/2]*(1/a)*[e^[x/a] + e^[-x/a]]
= [1/2]*[e^[x/a] + e^[-x/a]]

Answer 2

y = a/2 (e^x / a - (e^(-x)) / a))

You can factor out the 1/a in the inner parentheses.

y = a/2 (1/a) (e^x - (e^(-x)))

(a/2) * (1/a) is .5, so...

y = .5 (e^x - e^(-x))

I'm going to distribute first, just because I don't like parentheses.

y = .5e^x - .5e^(-x)

e^x derives to itself, and e^(-x) derives to -e^(-x). (Right?)

So...

dy/dx = .5e^x + .5e^(-x)

It's a fairly straightforward thing once you get the a out, though anything involving e can get a little tricky.

Hope it helped.