Continuity and Differentiability — Class 12 Maths Solution

therefore,${f^\prime }\left( {\frac{\pi }{3}} \right) = \frac{{\sqrt 3 + 1}}{2}$

We know that, $\frac{\pi }{4} < x < \frac{\pi }{2},\sin x > \cos x$
therefore,$\cos x - \sin x \le 0$ i.e., $f(x) = - (\cos x - \sin x)$

${f^\prime }(x) = - [ - \sin x - \cos x]$

therefore,${f^\prime }\left( {\frac{\pi }{3}} \right) = - \left( {\frac{{ - \sqrt 3 }}{2} - \frac{1}{2}} \right) = \left( {\frac{{\sqrt 3 + 1}}{2}} \right)$