Answer:
-0.032
Step-by-step explanation:
The given equation is
[tex] {x}^{3} + 5x + y = {y}^{3}[/tex]
We differentiate implicitly to get:
[tex]3{x}^{2} + 5 + \frac{dy}{dx} = 3 {y}^{2} \frac{dy}{dx} [/tex]
[tex] \frac{dy}{dx} - 3 {y}^{2} \frac{dy}{dx} = - ( 3{x}^{2} + 5)[/tex]
[tex](1 - 3 {y}^{2} )\frac{dy}{dx} = - ( 3{x}^{2} + 5)[/tex]
[tex] \frac{dy}{dx} = \frac{ - ( 3{x}^{2} + 5)}{1 - 3 {y}^{2} } [/tex]
[tex]\frac{dy}{dx} = \frac{ 3{x}^{2} + 5}{ 3 {y}^{2} - 1 } [/tex]
We differentiate again using the quotient rule:
[tex]\frac{d^{2} y}{d {x}^{2} } = \frac{ 6 {x} (3 {y}^{2} - 1) - (3 {x}^{2} + 5)(6y \frac{dy}{dx}) }{ (3 {y}^{2} - 1)^{2} } [/tex]
At (1,2), x=1, and y=2
[tex]\frac{d^{2} y}{d {x}^{2} } = \frac{ 6 { \times 1} (3 { \times 2}^{2} - 1) - (3 { \times 1}^{2} + 5)(6 \times 2 \times \frac{8}{11} ) }{ (3 { \times 2}^{2} - 1)^{2} } [/tex]
[tex]\frac{d^{2} y}{d {x}^{2} } = - 0.032[/tex]
