Question

Find dy/dx if √x+√y=√a

Answer 1

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

Given expression is

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

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

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

We know,

$\boxed{\tt{ \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v}}$

and

$\boxed{\tt{ \dfrac{d}{dx}k \: = \: 0 \: }}$

So, using these results, we get

$\rm \: \dfrac{d}{dx} \sqrt{x} + \dfrac{d}{dx} \sqrt{y} = 0$

We know,

$\boxed{\tt{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } \: }}$

So, using this result, we get

$\rm \: \dfrac{1}{2 \sqrt{x} } + \dfrac{1}{2 \sqrt{y} } \dfrac{dy}{dx} = 0$

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

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

$\rm\implies \:\boxed{\tt{ \: \: \dfrac{dy}{dx} = - \: \frac{ \sqrt{y} }{ \sqrt{x} } \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

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