Let y=y(x) be a function of x satisfying y√1-x^2=k-x√1-y^2 where k is a constant and y( 1 2)=- 1 4.Then dy dx at x= 1 2 , is equal to

$$x=\frac{1}{2},y=\frac{-1}{4}\Rightarrow xy=\frac{-1}{8}$$x=21​,y=4−1​⇒xy=8−1​

$$y.\frac{(-2x)}{2\sqrt{1-{x}^{2}}}+{y}^{'}.\sqrt{1-{x}^{2}}=0-{1.\sqrt{1-{y}^{2}}+\frac{x.(-2y)}{2\sqrt{1-{y}^{2}}}{y}^{'}}$$y.21−x2​(−2x)​+y′.1−x2​=0−1.1−y2​+21−y2​x.(−2y)​y′

$$-\frac{xy}{\sqrt{1-{x}^{2}}}+{y}^{'}\sqrt{1-{x}^{2}}=-\sqrt{1-{y}^{2}}+\frac{xy.{y}^{'}}{\sqrt{1-{y}^{2}}}$$−1−x2​xy​+y′1−x2​=−1−y2​+1−y2​xy.y′​

$${y}^{'}(\sqrt{1-{x}^{2}}-\frac{xy}{\sqrt{1-{y}^{2}}})=\frac{xy}{\sqrt{1-{x}^{2}}}-\sqrt{1-{y}^{2}}$$y′(1−x2​−1−y2​xy​)=1−x2​xy​−1−y2​

$${y}^{'}(\frac{\sqrt{3}}{2}+\frac{1}{8.\frac{\sqrt{15}}{4}})=\frac{-1}{8.\sqrt{\frac{3}{4}}}-\frac{\sqrt{15}}{4}$$y′(23​​+8.415​​1​)=8.43​​−1​−415​​

$${y}^{'}(\frac{\sqrt{45}+1}{2\sqrt{15}})=-\frac{(1+\sqrt{45})}{4\sqrt{3}}$$y′(215​45​+1​)=−43​(1+45​)​

$${y}^{'}=-\frac{\sqrt{15}}{2\sqrt{3}}=-\frac{\sqrt{5}}{2}$$y′=−23​15​​=−25​​