Continuity and Differentiability — Class 12 Maths Solution

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

$\Rightarrow$ $\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt y }}\frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} = - \frac{{\sqrt y }}{{\sqrt x }}$
therefore,${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{4},\frac{1}{4}} \right)}} = \frac{{ - \frac{1}{2}}}{{\frac{1}{2}}} = - 1$