If f(x) = sqrt{1 + cos^2(x^2)},then f^{prime} | vedclass

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(B) Given $f(x) = \sqrt{1 + \cos^2(x^2)}$.Using the chain rule,let $u = 1 + \cos^2(x^2)$,so $f(x) = \sqrt{u}$.$f^{\prime}(x) = \frac{1}{2\sqrt{u}} \cd

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$-\sqrt{\frac{\pi}{6}}$

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(B) Given $f(x) = \sqrt{1 + \cos^2(x^2)}$.Using the chain rule,let $u = 1 + \cos^2(x^2)$,so $f(x) = \sqrt{u}$.$f^{\prime}(x) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{1 + \cos^2(x^2)}} \cdot \frac{d}{dx}(1 + \cos^2(x^2))$.Now,$\frac{d}{dx}(\cos^2(x^2)) = 2\cos(x^2) \cdot (-\sin(x^2)) \cdot 2x = -2x \sin(2x^2)$.Thus,$f^{\prime}(x) = \frac{-2x \sin(2x^2)}{2\sqrt{1 + \cos^2(x^2)}} = \frac{-x \sin(2x^2)}{\sqrt{1 + \cos^2(x^2)}}$.At $x = \frac{\sqrt{\pi}}{2}$,$x^2 = \frac{\pi}{4}$.$f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right) = \frac{-\frac{\sqrt{\pi}}{2} \sin(2 \cdot \frac{\pi}{4})}{\sqrt{1 + \cos^2(\frac{\pi}{4})}} = \frac{-\frac{\sqrt{\pi}}{2} \sin(\frac{\pi}{2})}{\sqrt{1 + (\frac{1}{\sqrt{2}})^2}} = \frac{-\frac{\sqrt{\pi}}{2} \cdot 1}{\sqrt{1 + \frac{1}{2}}} = \frac{-\frac{\sqrt{\pi}}{2}}{\sqrt{\frac{3}{2}}} = -\frac{\sqrt{\pi}}{2} \cdot \sqrt{\frac{2}{3}} = -\frac{\sqrt{\pi}}{\sqrt{2} \cdot \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{\pi}}{\sqrt{6}} = -\sqrt{\frac{\pi}{6}}$.

If $f(x) = \sqrt{1 + \cos^2(x^2)}$,then $f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)$ is

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