Question

For the curve $\sqrt{x}+\sqrt{y}=1,\frac{dy}{dx}\mathrm{at}\left(1/4,1/4\right)\mathrm{is}$
(a) 1/2
(b) 1
(c) −1
(d) 2

Answer 1

(c) −1

$$\begin{gathered} \mathrm{We}\mathrm{have},\sqrt{x}+\sqrt{y}=1 \\ \mathrm{Differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get}, \\ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0 \\ \Rightarrow \frac{1}{2\sqrt{y}}\frac{dy}{dx}=-\frac{1}{2\sqrt{x}} \\ \Rightarrow \frac{dy}{dx}=-\frac{1}{2\sqrt{x}}\times \frac{2\sqrt{y}}{1} \\ \Rightarrow \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}} \\ \mathrm{Now},{\left[\frac{dy}{dx}\right]}_{\left(\frac{1}{4},\frac{1}{4}\right)}=-\frac{\sqrt{{\displaystyle \frac{1}{4}}}}{\sqrt{{\displaystyle \frac{1}{4}}}}=-1 \end{gathered}$$