Question

if x=y√1-y²,then dy/dx who answer it I will make them brainlist​

Answer 1

Answer:

Simplyfy

$x = y \sqrt{1 - {y}^{2} } \\ {x}^{2} = {y}^{2} (1 - {y}^{2} ) \\ \\ let \: \: {y}^{2} = u \\ \\ {x}^{2} = u(1 - u) \\ \\ {u}^{2} - u + {x}^{2} = 0 \\ \\ shreedhraharya \: \: formula\\ u = \frac{1 \pm \sqrt{1 - 4 {x}^{2} } }{2} = {y}^{2} \\ \\ y = \sqrt{ \frac{1 \pm \sqrt{1 - 4 {x}^{2} } }{2} } \\ \\ \frac{dy}{dx} = \frac{1}{2} \bigg( \frac{1 \pm \sqrt{1 - 4 {x}^{2} } }{2} \bigg) {}^{ - \frac{1}{2} } .( \pm \frac{4x}{ \sqrt{1 - 4 {x}^{2} } } \\ \\if\: \sqrt{1-4x^2}=k \\\\ \frac{dy}{dx}= \pm\frac{ \sqrt2x}{k \sqrt{\pm \:k+1}}$