Help...y=ln((e^theta)/1+e^theta)?

Question

BTW, that's e to the power of theta, just in case. I don't know how to type those on my compooty.

I'd like to know how to arrive at the answer, which is in my book. Calc II is not really my bag.

I need the derivative. Thanks

Hey thanks. But I'm still a little confused. In line 5, the u' is the same as du correct? But where did the
u'/u(1+u) come from?

Answer 1

Let u= e^theta.

Then y = ln(u/(1+u))

Now, ln(a/b) = ln(a) - ln(b)

So y = ln(u) - ln(1+u)

So y' = u' (1/u - 1/(1+u)) = u' /[u(u+1)]

But u' = e^theta = u, so:

y' = u/[u(u+1)] = 1/(u+1) = 1/(1+e^theta)

Answer 2

So, y=ln(e^theta/1+e^theta)
Using the quotient rule of law of logs, you get y = ln(e^theta) - ln(1+e^theta)
Now you can derive it.
y'=1/e^theta*e^theta-1/1+e^theta*1/1+e^theta
Now simplify.
=1-e^theta/1+e^theta
=1+e^theta-e^theta/1+e^theta
=1/1+e^theta