$a={{{f}'}_{+}}(\frac{\pi }{4})=\underset{x\to {{(\frac{\pi }{4})}^{+}}}{\mathop{\lim }}\,\frac{f(x)-f(\frac{\pi }{4})}{x-\frac{\pi }{4}}$

$=\underset{x\to {{(\frac{\pi }{4})}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{1-\sin 2x}-0}{x-\frac{\pi }{4}}=\underset{x\to {{(\frac{\pi }{4})}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x}}{x-\frac{\pi }{4}}=\underset{x\to {{(\frac{\pi }{4})}^{+}}}{\mathop{\lim }}\,\frac{\left| \sin x-\cos x \right|}{x-\frac{\pi }{4}}$

$=\underset{x\to {{(\frac{\pi }{4})}^{+}}}{\mathop{\lim }}\,\frac{\left| \sqrt{2}\sin (x-\frac{\pi }{4}) \right|}{(x-\frac{\pi }{4})}=\underset{x\to {{(\frac{\pi }{4})}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{2}\sin (x-\frac{\pi }{4})}{x-\frac{\pi }{4}}=\sqrt{2}\Rightarrow a=\sqrt{2}$ 

حال مشتق چپ تابع را در $x=\frac{\pi }{4}$ به‌دست می‌آوریم:

$b={{{f}'}_{-}}(\frac{\pi }{4})=\underset{x\to {{(\frac{\pi }{4})}^{-}}}{\mathop{\lim }}\,\frac{\left| \sin x-\cos x \right|}{x-\frac{\pi }{4}}$

$=\underset{x\to {{(\frac{\pi }{4})}^{-}}}{\mathop{\lim }}\,\frac{\left| \sqrt{2}\sin (x-\frac{\pi }{4}) \right|}{x-\frac{\pi }{4}}=\underset{x\to {{(\frac{\pi }{4})}^{-}}}{\mathop{\lim }}\,\frac{-\sqrt{2}\sin (x-\frac{\pi }{4})}{x-\frac{\pi }{4}}$

$=-\sqrt{2}\Rightarrow b=-\sqrt{2}\Rightarrow a+2b=\sqrt{2}-2\sqrt{2}=-\sqrt{2}$