If $x \cos (k+y)=\cos y$, then $\frac{d y}{d x}$ at $y=\frac{\pi}{2}$ is

If $x=a(\cos \theta+\theta \sin \theta), y=f(\theta), f(2 \pi)=0$, $\frac{d y}{d x}=\frac{\tan \theta}{\theta}, \theta \neq 0$ and $\theta \neq(2 n+1) \frac{\pi}{2}$, then $f\left(\frac{\pi}{3}\right)=$

If $a f(x)+b f\left(\frac{1}{x}\right)=x+1$, and $\frac{d}{d x}\left(x^2 f(x)\right)=2 x^2+2 x+\frac{1}{3}$, then $a-b$

If $f(x)=\sin \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right)$, then $f^{\prime}(1)=$