If {x^2} + {y^2} = 1,then find the relation b | vedclass

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(B) Given the equation: ${x^2} + {y^2} = 1$. Differentiating both sides with respect to $x$: $\frac{d}{dx}({x^2} + {y^2}) = \frac{d}{dx}(1)$ $2x + 2y

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$yy'' + (y')^2 + 1 = 0$

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(B) Given the equation: ${x^2} + {y^2} = 1$. Differentiating both sides with respect to $x$: $\frac{d}{dx}({x^2} + {y^2}) = \frac{d}{dx}(1)$ $2x + 2y \frac{dy}{dx} = 0$ $x + y y' = 0$ Differentiating again with respect to $x$ using the product rule on $y y'$: $\frac{d}{dx}(x) + \frac{d}{dx}(y y') = 0$ $1 + (y \cdot y'' + y' \cdot y') = 0$ $1 + y y'' + (y')^2 = 0$ Thus,the relation is $yy'' + (y')^2 + 1 = 0$.

If ${x^2} + {y^2} = 1$,then find the relation between $y'$ and $y''$,where $y' = \frac{dy}{dx}$ and $y'' = \frac{d^2y}{dx^2}$.

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