If-f-x-sin-cos-x-pi-x-f-pi-2-

Question Number 187695 by mnjuly1970 last updated on 20/Feb/23

I f, f (x) = \frac{s i \underset{}{n} (c o s (x))}{\overset{}{} \sqrt{\frac{π}{x}}}

\Rightarrow f^{'} (\frac{π}{2}) = ?

Answered by horsebrand11 last updated on 20/Feb/23

f (x) = \frac{\sqrt{x} \sin (\cos x)}{\sqrt{π}}

f^{'} (\frac{π}{2}) = \lim_{x \to \frac{π}{2}} \frac{f (x) - f (\frac{π}{2})}{x - \frac{π}{2}}

= \frac{1}{\sqrt{π}} \lim_{x \to \frac{π}{2}} \frac{\sqrt{x} \sin (\cos x) - \sqrt{\frac{π}{2}} \sin (\cos \frac{π}{2})}{x - \frac{π}{2}}

= \frac{1}{\sqrt{π}} . \frac{\sqrt{π}}{\sqrt{2}} . \lim_{x \to \frac{π}{2}} \frac{\sin (\cos x)}{x - \frac{π}{2}}

= \frac{1}{\sqrt{2}} . \lim_{x \to 0} \frac{\sin (- \sin x)}{x}

= \frac{1}{\sqrt{2}} . \lim_{x \to 0} \frac{- \sin x}{x} = - \frac{1}{\sqrt{2}}

Answered by cortano12 last updated on 20/Feb/23

f (x) = \frac{\sqrt{x} \sin (\cos x)}{\sqrt{π}}

f^{'} (x) = \frac{1}{\sqrt{π}} [\frac{\sin (\cos x)}{2 \sqrt{x}} - \sqrt{x} \sin x \cos (\cos x)]

f^{'} (\frac{π}{2}) = \frac{1}{\sqrt{π}} [0 - \frac{\sqrt{π}}{\sqrt{2}} . 1] = - \frac{\sqrt{2}}{2}