Answer
$$y'=\frac{1}{1+e^{\theta}}$$
Work Step by Step
$$y=\ln\Big(\frac{e^\theta}{1+e^{\theta}}\Big)$$
We have $$(\ln \theta)'=\frac{1}{\theta}$$ $$(e^{\theta})'=e^\theta$$
So the derivative of $y$ is $$y'=\frac{1}{\frac{e^{\theta}}{1+e^\theta}}\times\Big(\frac{e^\theta}{1+e^{\theta}}\Big)'=\frac{1+e^{\theta}}{e^{\theta}}\times\frac{e^{\theta}(1+e^\theta)-e^{\theta}(0+e^\theta)}{(1+e^{\theta})^2}$$
$$y'=\frac{1+e^{\theta}}{e^{\theta}}\times\frac{e^\theta+e^{2\theta}-e^{2\theta}}{(1+e^{\theta})^2}=\frac{1+e^{\theta}}{e^{\theta}}\times\frac{e^\theta}{(1+e^{\theta})^2}$$
$$y'=\frac{1}{1+e^{\theta}}$$
