(a) If xy = e^x - y, then find dy/dx.

$\displaystyle \Rightarrow \left(x+e^{x-y}\right) \dfrac{dy}{dx}=e^{x-y}-y \Rightarrow \dfrac{dy}{dx}=\dfrac{e^{x-y}-y}{x+e^{x-y}}$ or $\displaystyle \dfrac{y(x-1)}{x(y+1)}$

$\displaystyle x y=e^{x-y} \Rightarrow \log x+\log y=(x-y)$

$\displaystyle \Rightarrow \dfrac{1}{x}+\dfrac{1}{y} \dfrac{dy}{dx}=1-\dfrac{dy}{dx}$

$\displaystyle \Rightarrow \dfrac{dy}{dx}=\dfrac{1-\dfrac{1}{x}}{\dfrac{1}{y}+1}=\dfrac{y(x-1)}{x(y+1)}$

$\displaystyle y=\tan^{-1}\left(\dfrac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}\right)$

$\displaystyle y=\tan^{-1}\left(\dfrac{\cos\left(\dfrac{\theta}{2}\right)-\sin\left(\dfrac{\theta}{2}\right)}{\cos\left(\dfrac{\theta}{2}\right)+\sin\left(\dfrac{\theta}{2}\right)}\right)=\tan^{-1}\left(\tan\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\right)$

Thus, $\displaystyle y=\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right) \Rightarrow \dfrac{dy}{d\theta}=\dfrac{-1}{2}$