${\cos ^{ - 1}}\left( {\frac{{\sin x + \cos x}}{{\sqrt 2 }}} \right), - \frac{\pi }{4} < x < \frac{\pi }{4}$

Let $y = {\cos ^{ - 1}}\left( {\frac{{\sin x + \cos x}}{{\sqrt 2 }}} \right)$

therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\cos ^{ - 1}}\left( {\frac{{\sin x + \cos x}}{{\sqrt 2 }}} \right)$

$= \frac{{ - 1}}{{\sqrt {1 - {{\left( {\frac{{\sin x + \cos x}}{{\sqrt 2 }}} \right)}^2}} }} \cdot \frac{d}{{dx}}\left( {\frac{{\sin x + \cos x}}{{\sqrt 2 }}} \right)$

$= \frac{{ - 1}}{{\sqrt {4 - \frac{{\left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x \cdot \cos x} \right)}}{2}} \cdot \frac{1}{{\sqrt 2 }}(\cos x - \sin x)}}$

$= \frac{{ - 1 \cdot \sqrt 2 }}{{\sqrt {1 - \sin 2x} }} \cdot \frac{1}{{\sqrt 2 }}(\cos x - \sin x)$

$= \frac{{ - 1(\cos x - \sin x)}}{{(\cos x - \sin x)}} = - 1$

$\frac{dy}{dx} = = \frac{{ - 1(\cos x - \sin x)}}{{(\cos x - \sin x)}} = - 1$