Question

For the curve $\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=1,$find $\frac{\mathrm{dy}}{dx}$at $\left(\frac{1}{4},\frac{1}{4}\right)$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $-1$
• D. $2$

Answer 1

The correct option is C

$-1$

Explanation for the correct option:

Step1. Differentiate with respect to $x$:

Given $\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=1,$

Differentiate the function with respect to $x$

$\frac{1}{2\sqrt{x}}+\left(\frac{1}{2\sqrt{y}}\right)\frac{dy}{dx}=0$ $\left[\because \frac{d\left(\sqrt{x}\right)}{dx}=\frac{1}{2\sqrt{x}}\right]$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{\left({\displaystyle \frac{-1}{2\sqrt{x}}}\right)}{\left({\displaystyle \frac{1}{2\sqrt{y}}}\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{-\sqrt{x}}{\sqrt{y}} \end{gathered}$$

Step2. Find the value of $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{}\mathbf{at}\mathbf{}\left(\frac{1}{4},\frac{1}{4}\right)$:

Put $x=\frac{1}{4}$and $y=\frac{1}{4}$, then

$\begin{array}{rcl}{\frac{dy}{dx}}_{\left(\frac{1}{4},\frac{1}{4}\right)}& =& \frac{-1}{{\displaystyle \frac{\sqrt{4}}{{\displaystyle \frac{1}{\sqrt{4}}}}}}\\ & =& -1\end{array}$

Hence, Option (C) is the correct answer.