Maclaurin series of $y=\ln (\dfrac{1+e^{-x}}{2})$ with $\dfrac{dy}{dx}=\dfrac{e^{-y}}{2}-1$
This question requires you to use the given result of $dy/dx$
I've worked out that $y''=\dfrac{-e^{-y}}{2}\dfrac{dy}{dx}$ (is this right?)
Then using the product rule: $y'''=\dfrac{-e^{-y}}{2}\dfrac{d^2y}{dx^2}+\dfrac{e^{-y}}{2}\dfrac{dy}{dx}$
When I plug in $ x=0 $ I do not get the required answer - for the functions that have $dy/dx$ or $d^2y/dx^2$ in them can I substitute in the value I calculated for $dy/dx$ etc. when I substituted $x=0$ in? Where am I going wrong?
It seems a little confusing your question to me..so you want to obtain the Maclaurin expansion given that you know that $y\prime = \frac{\mathrm{e}^{-y}}{2}-1$? but in any case your calculation of $y''$ is correct. however for $y'''$ we have $$ y''' = \dfrac{d}{dx}\left(-\frac{\mathrm{e}^{-y}}{2}y'\right) = -\frac{\mathrm{e}^{-y}}{2}y'' +\frac{\mathrm{e}^{-y}}{2}(y')^2\neq -\frac{\mathrm{e}^{-y}}{2}y'' +\frac{\mathrm{e}^{-y}}{2}(y') $$ with the latter being what you have.
now we have $$ y''' = -\frac{\mathrm{e}^{-y}}{2}y'' +\frac{\mathrm{e}^{-y}}{2}(y')^2 =-\frac{\mathrm{e}^{-y}}{2}\left(-\frac{\mathrm{e}^{-y}}{2}y'\right)+\frac{\mathrm{e}^{-y}}{2}\left(\frac{\mathrm{e}^{-y}}{2}-1\right)y'\\ =\left[\frac{\mathrm{e}^{-2y}}{4}+\frac{\mathrm{e}^{-2y}}{4}-\frac{\mathrm{e}^{-y}}{2}\right]y' =\left[\frac{\mathrm{e}^{-2y}}{2}-\frac{\mathrm{e}^{-y}}{2}\right]y' $$
$\textbf{appendix}$ $$ \dfrac{d}{dx}\left(-\frac{\mathrm{e}^{-y}}{2}y'\right) = -\frac{1}{2}\left[y'\dfrac{d}{dx}\left(\mathrm{e}^{-y}\right)+\mathrm{e}^{-y}\dfrac{d}{dx}(y')\right] = -\frac{1}{2}\left[-y'\mathrm{e}^{-y}y'+\mathrm{e}^{-y}y''\right] = \frac{1}{2}\mathrm{e}^{-y}(y')^2+-\frac{1}{2}\mathrm{e}^{-y}y'' $$