Choose the correct alternative in the following:
Question:

Choose the correct alternative in the following:

For the curve $\cdot \sqrt{x}+\sqrt{y}=1, \frac{d y}{d x} \cdot$ at $(1 / 4,1 / 4)$ is

A. $1 / 2$

B. 1

C. $-1$

D. 2

Solution:

$\sqrt{x}+\sqrt{y}=1$

Differentiating w.r.t $\mathrm{x}$ we get,

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})+\frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{y}})=\frac{\mathrm{d}}{\mathrm{dx}}(1)$

$\Rightarrow \frac{1}{2 \sqrt{\mathrm{x}}}+\frac{1}{2 \sqrt{\mathrm{y}}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=0 \because \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\sqrt{\frac{\mathrm{y}}{\mathrm{x}}}$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(\mathrm{x}, \mathrm{y})=\left(\frac{1}{4} \frac{1}{4}\right)}=-\sqrt{\frac{\frac{1}{4}}{\frac{1}{4}}}=-1$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(\mathrm{x}, \mathrm{y})=\left(\frac{1}{4} \frac{1}{4}\right)}=-1=(\mathrm{C})$