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प्रश्न
\[\frac{2^x \cot x}{\sqrt{x}}\]
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उत्तर
\[\frac{2^x \cot x}{\sqrt{x}} = 2^x \cot x \left( x^\frac{- 1}{2} \right)\]
\[\text{ Let } u = 2^x ; v = \cot x; w = x^\frac{- 1}{2} \]
\[\text{ Then }, u' = 2^x \log 2; v' = - {cosec}^2 x; w' = \frac{- 1}{2} x^\frac{- 3}{2} \]
\[\text{ Using the product rule }:\]
\[\frac{d}{dx}\left( uvw \right) = u'vw + uv'w + uvw'\]
\[\frac{d}{dx}\left[ 2^x \cot x \left( x^\frac{- 1}{2} \right) \right] = 2^x \log 2 . \cot x . x^\frac{- 1}{2} + 2^x \left( - {cosec}^2 x \right) x^\frac{- 1}{2} + 2^x \cot x\left( \frac{- 1}{2} x^\frac{- 3}{2} \right)\]
\[ = 2^x \log 2 . \cot x . \frac{1}{\sqrt{x}} + 2^x \left( - {cosec}^2 x \right)\frac{1}{\sqrt{x}} + 2^x \cot x\left( \frac{- 1}{2x\sqrt{x}} \right)\]
\[ = \frac{2^x}{\sqrt{x}}\left( \log 2 . \cot x - {cosec}^2 x - \frac{\cot x}{2x} \right)\]
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