If x^2+xy+y^2=k, then frac{d^2y}{dx^2}= | vedclass

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(A) Given the equation $x^2+xy+y^2=k$. Differentiating both sides with respect to $x$: $2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$ $(x+2y)\frac{dy

Vedclass · Accepted Answer

$\frac{-6k}{(x+2y)^3}$

Vedclass · Answer

(A) Given the equation $x^2+xy+y^2=k$. Differentiating both sides with respect to $x$: $2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$ $(x+2y)\frac{dy}{dx} = -(2x+y)$ $\frac{dy}{dx} = -\frac{2x+y}{x+2y}$ Now, differentiating again with respect to $x$ using the quotient rule: $\frac{d^2y}{dx^2} = -\frac{(x+2y)\frac{d}{dx}(2x+y) - (2x+y)\frac{d}{dx}(x+2y)}{(x+2y)^2}$ $\frac{d^2y}{dx^2} = -\frac{(x+2y)(2+\frac{dy}{dx}) - (2x+y)(1+2\frac{dy}{dx})}{(x+2y)^2}$ Substituting $\frac{dy}{dx} = -\frac{2x+y}{x+2y}$: $\frac{d^2y}{dx^2} = -\frac{(x+2y)(2-\frac{2x+y}{x+2y}) - (2x+y)(1-\frac{4x+2y}{x+2y})}{(x+2y)^2}$ $\frac{d^2y}{dx^2} = -\frac{(x+2y)(\frac{2x+4y-2x-y}{x+2y}) - (2x+y)(\frac{x+2y-4x-2y}{x+2y})}{(x+2y)^2}$ $\frac{d^2y}{dx^2} = -\frac{3y - (2x+y)(\frac{-3x}{x+2y})}{(x+2y)^2} = -\frac{3y(x+2y) + 3x(2x+y)}{(x+2y)^3}$ $\frac{d^2y}{dx^2} = -\frac{3xy+6y^2+6x^2+3xy}{(x+2y)^3} = -\frac{6(x^2+xy+y^2)}{(x+2y)^3}$ Since $x^2+xy+y^2=k$, we get: $\frac{d^2y}{dx^2} = \frac{-6k}{(x+2y)^3}$

If $x^2+xy+y^2=k$, then $\frac{d^2y}{dx^2}=$

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