Question

If
${x}^{2} + {y}^{2} = 1$
then
$\frac{ {d}^{2} \times x}{d {y}^{2} } =$
​

Answer 1

${x}^{2} + {y}^{2} = 1$

differentiating with respect with y

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

$\frac{dx}{dy} = \frac{ - y}{x}$

differentiating again with respect to y

$\frac{ {d}^{2} x}{d {y}^{2} } = \frac{ - 1x - ( \frac{dx}{dy})( - y) }{ {x}^{2} }$

put the value of dy/de in above expression.

$\frac{ {d}^{2}x }{dy} = \frac{ - y - \frac{( - y)}{x}( - y) }{ {x}^{2} }$

$\frac{ {d}^{2}x}{d {y}^{2} } = \frac{ - xy - {y}^{2} }{ {x}^{3} }$

Answer 2

Step-by-step explanation:

hope it helps you good.....