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

Explanation:

Step1: Differentiate (x^{3}+y^{3}=19) implicitly

Differentiate both sides with respect to (x): (3x^{2}+3y^{2}\frac{dy}{dx}=0) Solve for (\frac{dy}{dx}): (\frac{dy}{dx}=-\frac{x^{2}}{y^{2}})

Step2: Differentiate (\frac{dy}{dx}) to find (\frac{d^{2}y}{dx^{2}})

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}) (\frac{d^{2}y}{dx^{2}}=\frac{-2x\cdot y^{2}-(-x^{2})\cdot2y\frac{dy}{dx}}{y^{4}}) Substitute (\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}) into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-2xy^{2}+2x^{2}y\cdot\frac{x^{2}}{y^{2}}}{y^{4}}\ &=\frac{-2xy^{3}+2x^{4}}{y^{5}} \end{align*} ]

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

[ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-2(-2)(3)^{3}+2(-2)^{4}}{3^{5}}\ &=\frac{4\times27 + 2\times16}{243}\ &=\frac{108 + 32}{243}\ &=\frac{140}{243} \end{align*} ]

Answer:

(\frac{140}{243})