यदि f(x) = sqrt{cos^{-1} sqrt{1-x^2}} है, तो | vedclass

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(A) दिया गया है $f(x) = \sqrt{\cos^{-1} \sqrt{1-x^2}}$.मान लीजिए $x = \sin 	heta$, तो $\sqrt{1-x^2} = \cos 	heta$.अतः, $f(x) = \sqrt{\cos^{-1}(\cos

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$\sqrt{\frac{2}{\pi}}$

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(A) दिया गया है $f(x) = \sqrt{\cos^{-1} \sqrt{1-x^2}}$.मान लीजिए $x = \sin 	heta$, तो $\sqrt{1-x^2} = \cos 	heta$.अतः, $f(x) = \sqrt{\cos^{-1}(\cos 	heta)} = \sqrt{	heta} = \sqrt{\sin^{-1} x}$.अब, $x$ के सापेक्ष $f(x)$ का अवकलन करने पर:$f^{\prime}(x) = \frac{d}{dx}(\sin^{-1} x)^{1/2} = \frac{1}{2}(\sin^{-1} x)^{-1/2} \cdot \frac{d}{dx}(\sin^{-1} x) = \frac{1}{2\sqrt{\sin^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}}$.$x = \frac{1}{2}$ पर, $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ और $\sqrt{1-(\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.$f^{\prime}(\frac{1}{2}) = \frac{1}{2\sqrt{\pi/6}} \cdot \frac{1}{\sqrt{3}/2} = \frac{1}{\sqrt{\pi/6} \cdot \sqrt{3}} = \frac{1}{\sqrt{\pi/2}} = \sqrt{\frac{2}{\pi}}$.

यदि $f(x) = \sqrt{\cos^{-1} \sqrt{1-x^2}}$ है, तो $f^{\prime}\left(\frac{1}{2}\right) = $

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