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If π 2 Then Find D D X ( √ 1 + Cos 2 X 2 )

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प्रश्न

If \[\frac{\pi}{2}\] then find \[\frac{d}{dx}\left( \sqrt{\frac{1 + \cos 2x}{2}} \right)\]

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उत्तर

\[\sqrt{\frac{1 + \cos 2x}{2}}\]
\[ = \sqrt{\frac{2 \cos^2 x}{2}}\]
\[ = \sqrt{\cos^2 x}\]
\[ = - \cos x (\because\frac{\pi}{2}<x<\pi)\]
\[\frac{d}{dx}\left( \sqrt{\frac{1 + \cos 2x}{2}} \right)\]
\[ = \frac{d}{dx}\left( - \cos x \right)\]
\[ = - \left( - \sin x \right)\]
\[ = \sin x\]

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अध्याय 30: Derivatives - Exercise 30.6 [पृष्ठ ४७]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 30 Derivatives
Exercise 30.6 | Q 4 | पृष्ठ ४७

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