If A = frac{2^x cot x}{sqrt{x}},then frac{dA} | vedclass

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(A) Given $A = \frac{2^x \cot x}{\sqrt{x}}$.Applying the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:$\frac{dA}{dx} = \frac{\s

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$\frac{2^{x-1} \{ -2x \csc^2 x + \cot x \cdot \log(\frac{4^x}{e}) \}}{x^{3/2}}$

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(A) Given $A = \frac{2^x \cot x}{\sqrt{x}}$.Applying the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:$\frac{dA}{dx} = \frac{\sqrt{x} \frac{d}{dx}(2^x \cot x) - 2^x \cot x \frac{d}{dx}(\sqrt{x})}{x}$$= \frac{\sqrt{x} [2^x \ln 2 \cot x - 2^x \csc^2 x] - 2^x \cot x \frac{1}{2\sqrt{x}}}{x}$$= \frac{2^x [\sqrt{x} \ln 2 \cot x - \sqrt{x} \csc^2 x - \frac{\cot x}{2\sqrt{x}}]}{x}$$= \frac{2^x [2x \ln 2 \cot x - 2x \csc^2 x - \cot x]}{2x^{3/2}}$$= \frac{2^x [2x \ln 2 \cot x - \cot x - 2x \csc^2 x]}{2x^{3/2}}$$= \frac{2^{x-1} [\cot x (2x \ln 2 - 1) - 2x \csc^2 x]}{x^{3/2}}$Since $\ln(4^x/e) = \ln(4^x) - \ln e = x \ln 4 - 1 = 2x \ln 2 - 1$,the expression becomes:$= \frac{2^{x-1} \{ -2x \csc^2 x + \cot x \cdot \log(\frac{4^x}{e}) \}}{x^{3/2}}$.

If $A = \frac{2^x \cot x}{\sqrt{x}}$,then $\frac{dA}{dx} = $

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