Question

If $A=\frac{(2^{x}cotx)}{\sqrt{x}}$ then $\frac{dA}{dx}=$

• A. $\left(\frac{2^x}{\sqrt{x}}\right)[\log 2cotx–\cos e{c}^{2}x–\frac{(cotx)}{2x}]$
• B. $\left(\frac{2^x}{x}\right)[\log 2cotx+\cos e{c}^{2}x–\frac{(cotx)}{2x}]$
• C. $\left(\frac{2^x}{\sqrt{x}}\right)[\log 2cotx–\cos e{c}^{2}x–(cotx)]$
• D. None of these

Answer 1

The correct option is A

$\left(\frac{2^x}{\sqrt{x}}\right)[\log 2cotx–\cos e{c}^{2}x–\frac{(cotx)}{2x}]$

Finding the value of $\frac{\mathbf{d}\mathbf{A}}{\mathbf{d}\mathbf{x}}$:

Given, $A=\frac{(2^{x}cotx)}{\sqrt{x}}$

Differentiate with respect to $x$

$\begin{array}{rcl}\frac{dA}{dx}& =& \frac{\left[\sqrt{x}{\displaystyle \frac{d}{dx}}(2^{x}cotx)–(2^{x}cotx)\frac{d}{dx}(\sqrt{x})\right]}{{\left(\sqrt{x}\right)}^2}\end{array}$ $\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(\frac{u}{v}\right)\mathbf{=}\frac{\mathbf{u}\mathbf{v}\mathbf{\text{'}}\mathbf{-}\mathbf{v}\mathbf{u}\mathbf{\text{'}}}{{\mathbf{v}}^{\mathbf{2}}}\right]$

$=\frac{\left[\sqrt{x}\left({\displaystyle \frac{d}{dx}}{2}^x\right)cotx+2^x{\displaystyle \frac{d}{dx}}\left(cotx\right)-(2^{x}cotx)\left({\displaystyle \frac{1}{2\sqrt{x}}}\right)\right]}{x}$ $\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{uv}\mathbf{\right)}\mathbf{=}\mathbf{uv}\mathbf{\text{'}}\mathbf{+}\mathbf{vu}\right]$

$$\begin{gathered} =\frac{\left[\sqrt{x}(2^x\log 2cotx–2^x\cos e{c}^{2}x)-(2^{x}cotx)\left({\displaystyle \frac{1}{2\sqrt{x}}}\right)\right]}{x} \\ =\mathbf{\left(}\frac{{\mathbf{2}}^{\mathbf{x}}}{\sqrt{\mathbf{x}}}\mathbf{\right)}\mathbf{}\mathbf{[}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\frac{\mathbf{(}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{x}\mathbf{)}}{\mathbf{2}\mathbf{x}}\mathbf{}\mathbf{]} \end{gathered}$$

Hence, option (A) is the correct option.