Question

For the curve ${\displaystyle \sqrt{x}+\sqrt{y}}$ = 1, ${\displaystyle \frac{\text{dy}}{\text{dx}}}$ at ${\displaystyle (\frac{1}{4},\frac{1}{4})}$ is ______.

Answer 1

For the curve ${\displaystyle \sqrt{x}+\sqrt{y}}$ = 1, ${\displaystyle \frac{\text{dy}}{\text{dx}}}$ at ${\displaystyle (\frac{1}{4},\frac{1}{4})}$ is – 1.

Explanation:

Given that: ${\displaystyle \sqrt{x}+\sqrt{y}}$ = 1

Differentiating both sides w.r.t. x

${\displaystyle \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\cdot \frac{\text{dy}}{\text{dx}}}$ = 0

⇒ ${\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}} \frac{\text{dy}}{\text{dx}}}$ = 0

⇒ ${\displaystyle \frac{1}{\sqrt{y}}\frac{\text{dy}}{\text{dx}}=\frac{-1}{\sqrt{x}}}$

⇒ ${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{-\sqrt{y}}{\sqrt{x}}}$

∴ ${\displaystyle \frac{\text{dy}}{\text{dx}}}$ at ${\displaystyle (\frac{1}{4},\frac{1}{4})=-\frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}}$

= – 1.