Answer
$$y'=\frac{1}{\theta+1}-e^\theta$$
Work Step by Step
$$y=\ln(\theta+1)-e^{\theta}$$
Recall the following Derivative Rules: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$ $$\frac{d}{dx}(e^{x})=e^x$$
Therefore, we have $$y'=\Big(\ln(\theta+1)-e^{\theta}\Big)'=\frac{1}{\theta+1}(\theta+1)'-e^{\theta}$$ $$y'=\frac{1}{\theta+1}-e^\theta$$
