How would I take the derivative of this with respect to $\eta$?
$$ f = \frac{1}{2}\left[ -3 + \frac{27}{(2+x)^2} + \frac{1-\frac{1}{[1+(x-1)C(\eta)]^2}}{C(\eta)} + 2log(x) \right] $$
Since $x$ is not dependent on $\eta$, then 3 of the terms go away. So I would have:
$$ \frac{df}{d\eta} = \frac{d}{d\eta}\left[ \frac{1-\frac{1}{[1+(x-1)C(\eta)]^2}}{2C(\eta)} \right] $$
Now clearly the chain rule and power rule will be needed. I am getting somewhat confused however. Here is an attempt at the quotient rule:
$$ \frac{2C(\eta)\frac{d}{d\eta} \left( 1-\frac{1}{[1+(x-1)C(\eta)]^2} \right) - \left( 1-\frac{1}{[1+(x-1)C(\eta)]^2} \right)2C'(\eta) }{\left[ 2C(\eta) \right]^2} $$
Obviously, the $2's$ cancel:
$$ \frac{C(\eta)\frac{d}{d\eta} \left( 1-\frac{1}{[1+(x-1)C(\eta)]^2} \right) - \left( 1-\frac{1}{[1+(x-1)C(\eta)]^2} \right)C'(\eta) }{2\left[ C(\eta) \right]^2} $$
Now I am still left with the derivative you can see in the numerator. Isolating that, I have:
$$ \frac{d}{d\eta} \left( 1-\frac{1}{[1+(x-1)C(\eta)]^2} \right) $$
And the first term is 0, so we have:
$$ \frac{d}{d\eta} \left( \frac{-1}{[1+(x-1)C(\eta)]^2} \right) $$
Now
$$ \frac{\left([1+(x-1)C(\eta)]^2\right)(0) + \frac{d}{d\eta}[1+(x-1)C(\eta)]^2}{[1+(x-1)C(\eta)]^4} $$
Reducing:
$$ \frac{ \frac{d}{d\eta}[1+(x-1)C(\eta)]^2}{[1+(x-1)C(\eta)]^4} $$
So I have:
$$ \frac{2[1+(x-1)C(\eta)](x-1)C'(\eta)}{[1+(x-1)C(\eta)]^4} $$
Putting it back into the equation above:
$$ \frac{C(\eta)\frac{2[1+(x-1)C(\eta)](x-1)C'(\eta)}{[1+(x-1)C(\eta)]^4} - \left( 1-\frac{1}{[1+(x-1)C(\eta)]^2} \right)C'(\eta) }{2\left[ C(\eta) \right]^2} $$
does this look correct?
I would simplify $$ \frac{1-\frac{1}{[1+(x-1)C(\eta)]^2}}{C(\eta)} $$ to $$\frac{1}{C(\eta)}-\frac{1}{[1+(x-1)C(\eta)]^2C(\eta)}$$ before taking the derivative.