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If $x^2+y^2=1$ then $\left(y^{\prime}=\frac{d y}{d x}, y^{\prime}=\frac{d^2 y}{d x^2}\right)$
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The correct answer is:
$y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0$
Differentiating w.r.t. x, 2 x+2 y y^{\prime}=0$ or $x+y y^{\prime}=0$
Differentiating again w.r.t. $x, \quad 1+y^{\prime 2}+y y^{\prime \prime}=0$
Differentiating again w.r.t. $x, \quad 1+y^{\prime 2}+y y^{\prime \prime}=0$
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