if ( x^{3}+y^{3}=19 ), find the value of ( \frac{d^{2} y}{d x^{2}} ) at the point ( (-2,3) ).\nthe value of…

Explanation:

Step1: Differentiate implicitly for first - derivative

Differentiate (x^{3}+y^{3}=19) with respect to (x). Using the power rule ((u^{n})^\prime = nu^{n - 1}u^\prime), we have (3x^{2}+3y^{2}\frac{dy}{dx}=0). Solve for (\frac{dy}{dx}): [ \begin{align*} 3y^{2}\frac{dy}{dx}&=- 3x^{2}\ \frac{dy}{dx}&=-\frac{x^{2}}{y^{2}} \end{align*} ]

Step2: Differentiate (\frac{dy}{dx}) for second - derivative

Use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u=-x^{2}), (u^\prime=-2x), (v = y^{2}), (v^\prime = 2y\frac{dy}{dx}). [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{(-2x)y^{2}-(-x^{2})(2y\frac{dy}{dx})}{y^{4}}\ &=\frac{-2xy^{2}+2x^{2}y\frac{dy}{dx}}{y^{4}}\ &=\frac{-2x}{y^{2}}+\frac{2x^{2}\frac{dy}{dx}}{y^{3}} \end{align*} ] Substitute (\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}) into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-2x}{y^{2}}+\frac{2x^{2}(-\frac{x^{2}}{y^{2}})}{y^{3}}\ &=\frac{-2x}{y^{2}}-\frac{2x^{4}}{y^{5}} \end{align*} ]

Step3: Substitute (x=-2), (y = 3)

[ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-2(-2)}{3^{2}}-\frac{2(-2)^{4}}{3^{5}}\ &=\frac{4}{9}-\frac{32}{243}\ &=\frac{4\times27 - 32}{243}\ &=\frac{108 - 32}{243}\ &=\frac{76}{243} \end{align*} ]

Answer:

(\frac{76}{243})