Question

${(\sqrt{x}+\frac{1}{\sqrt{x}})}^5$ differentiate function w.r.t. $x$

Answer 1

$y={(\sqrt{x}+\frac{1}{\sqrt{x}})}^5$

$\Rightarrow y={\left[\frac{x+1}{\sqrt{x}}\right]}^5$

$\frac{dy}{dx}=5{\left(\frac{x+1}{\sqrt{x}}\right)}^{(5-1)}\times \frac{d}{dx}\left(\frac{x+1}{\sqrt{x}}\right)$

$=5\frac{(x+1{)}^4}{x^2}\times \left[\frac{\sqrt{x}\frac{d}{dx}(x+1)-(x+1)\frac{d}{dx}\sqrt{x}}{(\sqrt{x}{)}^2}\right]$

$=\frac{5(x+1{)}^4}{x^2}\left[\frac{\sqrt{x}-\left(\frac{x+1}{2\sqrt{x}}\right)}{x}\right]$

$\Rightarrow \frac{5}{x^3}(x+1{)}^4\left[\frac{2x-x-1}{2\sqrt{x}}\right]$

$=\frac{5}{x^3}(x+1{)}^4\left[\frac{2x-x-1}{2\sqrt{x}}\right]$

$\Rightarrow \frac{5}{2}\frac{(x+1{)}^4(x-1)}{x^3\sqrt{x}}$

$\Rightarrow \frac{5}{2}\frac{(x+1{)}^3}{x^3}\frac{(x^2-1)}{\sqrt{x}}$

$\frac{dy}{dx}\Rightarrow \frac{5}{2}{(1+\frac{1}{x})}^3\left(\frac{x^2-1}{\sqrt{x}}\right)$