Question

Find the derivative of $\frac{2^x\cot x}{\sqrt{x}}$

• A. $\frac{2^x}{\sqrt{x}}\{\log 2\cot x-{\csc }^{2}x-\frac{\cot x}{2x}\}$
• B. $\frac{2(x+1)}{\sqrt{x}}\{\log 2\cot x-{\csc }^{2}x-\frac{\cot x}{2x}\}$
• C. $\frac{2^x}{\sqrt{x}}\{\log 2\cot x-{\csc }^{2}x-\frac{\cot x}{x^2}\}$
• D. None of these

Answer 1

The correct option is A $\frac{2^x}{\sqrt{x}}\{\log 2\cot x-{\csc }^{2}x-\frac{\cot x}{2x}\}$

$\frac{d}{dx}\left(\frac{2^x\cot x}{\sqrt{x}}\right)$

$=\frac{\sqrt{x}\frac{d}{dx}(2^x.\cot x)-2^x\cot x\left(\frac{d}{dx}\sqrt{x}\right)}{{\left(\sqrt{x}\right)}^2}$

$=\frac{\sqrt{x}(-2^x{\csc }^{2}x+2^x\cot x.\log 2)-\frac{2^x\cot x}{2\sqrt{x}}}{{\left(\sqrt{x}\right)}^2}$

$=\frac{2x(-2^x{\csc }^{2}x+2^x\cot x.\log 2)-2^x\cot x}{2x\sqrt{x}}$

$=\frac{2^x(-2x{\csc }^{2}x+2x\cot x\log 2-\cot x)}{2x\sqrt{x}}$

$=\frac{2^x}{\sqrt{x}}(\frac{-2x{\csc }^{2}x}{2x}+\frac{2x\cot x\log 2}{2x}-\frac{\cot x}{2x})$

$=\frac{2^x}{\sqrt{x}}(-{\csc }^{2}x+\log 2\cot x-\frac{\cot x}{2x})$

$=\frac{2^x}{\sqrt{x}}\{\log 2\cot x-{\csc }^{2}x-\frac{\cot x}{2x}\}$