Question

\[\frac{2^x \cot x}{\sqrt{x}}\] 

Answer 1

\[\text{ Let } u = 2^x \cot x; v = \sqrt{x}\]
\[\text{ Then }, u' = - 2^x {cosec}^2 x + 2^x \log 2 \cot x; v' = \frac{1}{2\sqrt{x}}\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{2^x cot x}{\sqrt{x}} \right) = \frac{\sqrt{x}\left( - 2^x {cosec}^2 x + 2^x \log 2 \cot x \right) - 2^x \cot x\left( \frac{1}{2\sqrt{x}} \right)}{\left( \sqrt{x} \right)^2}\]
\[ = \frac{\frac{x\left( - 2^x {cosec}^2 x + 2^x \log 2 \cot x \right) - 2^{x - 1} \cot x}{\sqrt{x}}}{x}\]
\[ = \frac{2^x \left( - x {cosec}^2 x + x \cot x \log 2 - \left( \frac{1}{2} \right)\cot x \right)}{x\sqrt{x}}\]
\[ = \frac{2^x \left( - x {cosec}^2 x + x \cot x \log 2 - \left( \frac{1}{2} \right)\cot x \right)}{x^\frac{3}{2}}\]
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