If x^{2} y^{5}=(x+y)^{7},then frac{d^{2} y}{d | vedclass

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(D) Given,$x^{2} y^{5}=(x+y)^{7}$.Taking the natural logarithm on both sides,we get:$2 \ln x + 5 \ln y = 7 \ln (x+y)$.Differentiating both sides with

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(D) Given,$x^{2} y^{5}=(x+y)^{7}$.Taking the natural logarithm on both sides,we get:$2 \ln x + 5 \ln y = 7 \ln (x+y)$.Differentiating both sides with respect to $x$:$\frac{2}{x} + \frac{5}{y} \frac{dy}{dx} = \frac{7}{x+y} \left(1 + \frac{dy}{dx}\right)$.Rearranging the terms to solve for $\frac{dy}{dx}$:$\frac{dy}{dx} \left(\frac{5}{y} - \frac{7}{x+y}\right) = \frac{7}{x+y} - \frac{2}{x}$.$\frac{dy}{dx} \left(\frac{5x + 5y - 7y}{y(x+y)}\right) = \frac{7x - 2x - 2y}{x(x+y)}$.$\frac{dy}{dx} \left(\frac{5x - 2y}{y(x+y)}\right) = \frac{5x - 2y}{x(x+y)}$.Thus,$\frac{dy}{dx} = \frac{y}{x}$.Now,differentiating again with respect to $x$ using the quotient rule:$\frac{d^{2}y}{dx^{2}} = \frac{x \frac{dy}{dx} - y(1)}{x^{2}}$.Substituting $\frac{dy}{dx} = \frac{y}{x}$:$\frac{d^{2}y}{dx^{2}} = \frac{x(y/x) - y}{x^{2}} = \frac{y - y}{x^{2}} = 0$.

If $x^{2} y^{5}=(x+y)^{7}$,then $\frac{d^{2} y}{d x^{2}}$ is equal to

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