Question

y = √x/a +√a/x find dy/dx
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y = \sqrt{\dfrac{x}{a} } + \sqrt{\dfrac{a}{x} }$

can be rewritten as

$\rm \: y = \dfrac{ \sqrt{x} }{ \sqrt{a} } + \dfrac{ \sqrt{a} }{ \sqrt{x} }$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx} y = \dfrac{d}{dx} \bigg(\dfrac{ \sqrt{x} }{ \sqrt{a} } + \dfrac{ \sqrt{a} }{ \sqrt{x} }\bigg)$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{a} }\dfrac{d}{dx} \sqrt{x} + \sqrt{a} \dfrac{d}{dx} \dfrac{1}{ \sqrt{x} }$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{a} }\dfrac{d}{dx} {\bigg(x\bigg) }^{\dfrac{1}{2} } + \sqrt{a} \dfrac{d}{dx}{\bigg(x\bigg) }^{ - \dfrac{1}{2} }$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{a} }\dfrac{1}{2} {\bigg(x\bigg) }^{\dfrac{1}{2} - 1} - \dfrac{1}{2} \sqrt{a} {\bigg(x\bigg) }^{ - \dfrac{1}{2} - 1}$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{ 2\sqrt{a} } {\bigg(x\bigg) }^{ - \dfrac{1}{2}} - \dfrac{ \sqrt{a} }{2} {\bigg(x\bigg) }^{ - \dfrac{3}{2}}$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{ 2\sqrt{a} \sqrt{x} } - \dfrac{ \sqrt{a} }{2x \sqrt{x} }$

$\rm \: \dfrac{dy}{dx} = \dfrac{x - a}{ 2x\sqrt{a} \sqrt{x} }$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{dy}{dx} = \dfrac{x - a}{ 2x\sqrt{a} \sqrt{x} } \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$