$\mathrm{y}=\frac{1+\mathrm{sinx}-\mathrm{cosx}}{1+\mathrm{sinx}+\mathrm{cosx}},$ then ${\mathrm{y}}_{2=}$

$$\begin{gathered} \mathrm{y}=\frac{1+\sin \mathrm{x}-\cos \mathrm{x}}{1+\sin \mathrm{x}+\cos \mathrm{x}} \\ =\frac{2{\sin }^2{\displaystyle \frac{\mathrm{x}}{2}}+2\sin {\displaystyle \frac{\mathrm{x}}{2}}\cos {\displaystyle \frac{\mathrm{x}}{2}}}{2{\cos }^2{\displaystyle \frac{\mathrm{x}}{2}}+2\sin {\displaystyle \frac{\mathrm{x}}{2}}\cos {\displaystyle \frac{\mathrm{x}}{2}}} \\ =\frac{2\sin {\displaystyle \frac{\mathrm{x}}{2}}\left(\sin {\displaystyle \frac{\mathrm{x}}{2}}+\cos {\displaystyle \frac{\mathrm{x}}{2}}\right)}{2\cos {\displaystyle \frac{\mathrm{x}}{2}}\left(\cos {\displaystyle \frac{\mathrm{x}}{2}}+\sin {\displaystyle \frac{\mathrm{x}}{2}}\right)} \\ =\tan \frac{\mathrm{x}}{2} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}={\sec }^2\frac{\mathrm{x}}{2}\left(\frac{1}{2}\right) \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{1}{2}\left(2\sec \frac{\mathrm{x}}{2}\right) \left(\sec \frac{\mathrm{x}}{2}·\tan \frac{\mathrm{x}}{2}\right)\frac{1}{2} \\ =\frac{1}{2}{\sec }^2\frac{\mathrm{x}}{2}·\tan \frac{\mathrm{x}}{2} \end{gathered}$$