find ( y^{prime prime} ) by implicit differentiation.\n\n( 5 x^{3}-3 y^{3}=7 )\n( y^{prime prime}= )

Explanation:

Step1: Differentiate both sides with respect to (x)

Differentiate (5x^{3}-3y^{3}=7) term - by - term. Using the power rule ((x^{n})^\prime=nx^{n - 1}) and the chain rule ((y^{n})^\prime=ny^{n - 1}y^\prime). The derivative of (5x^{3}) with respect to (x) is (15x^{2}), the derivative of (-3y^{3}) with respect to (x) is (-9y^{2}y^\prime), and the derivative of the constant (7) is (0). So, (15x^{2}-9y^{2}y^\prime = 0). Solve for (y^\prime): [ \begin{align*} -9y^{2}y^\prime&=- 15x^{2}\ y^\prime&=\frac{15x^{2}}{9y^{2}}=\frac{5x^{2}}{3y^{2}} \end{align*} ]

Step2: Differentiate (y^\prime=\frac{5x^{2}}{3y^{2}}) with respect to (x) using the quotient rule

The quotient rule states that if (u = 5x^{2}) and (v = 3y^{2}), then ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). We know that (u^\prime = 10x) and (v^\prime=6yy^\prime). Substitute (u), (u^\prime), (v), and (v^\prime) into the quotient rule: [ \begin{align*} y^{\prime\prime}&=\frac{10x\cdot3y^{2}-5x^{2}\cdot6yy^\prime}{(3y^{2})^{2}}\ &=\frac{30xy^{2}-30x^{2}yy^\prime}{9y^{4}}\ \end{align*} ] Substitute (y^\prime=\frac{5x^{2}}{3y^{2}}) into the above - expression: [ \begin{align*} y^{\prime\prime}&=\frac{30xy^{2}-30x^{2}y\cdot\frac{5x^{2}}{3y^{2}}}{9y^{4}}\ &=\frac{30xy^{2}-\frac{150x^{4}}{3y}}{9y^{4}}\ &=\frac{30xy^{3}- 50x^{4}}{9y^{5}}\ &=\frac{10x(3y^{3}-5x^{3})}{9y^{5}} \end{align*} ] Since (5x^{3}-3y^{3}=7), then (3y^{3}-5x^{3}=-7). [ y^{\prime\prime}=\frac{10x(-7)}{9y^{5}}=-\frac{70x}{9y^{5}} ]

Answer:

(y^{\prime\prime}=-\frac{70x}{9y^{5}})