Question

Find the second derivative at the point (1,2), given the function below. x^3+5x+y=y^3 Round answer to 3 decimals.

Answer 1

Answer:

-0.032

Step-by-step explanation:

The given equation is

${x}^{3} + 5x + y = {y}^{3}$

We differentiate implicitly to get:

$3{x}^{2} + 5 + \frac{dy}{dx} = 3 {y}^{2} \frac{dy}{dx}$

$\frac{dy}{dx} - 3 {y}^{2} \frac{dy}{dx} = - ( 3{x}^{2} + 5)$

$(1 - 3 {y}^{2} )\frac{dy}{dx} = - ( 3{x}^{2} + 5)$

$\frac{dy}{dx} = \frac{ - ( 3{x}^{2} + 5)}{1 - 3 {y}^{2} }$

$\frac{dy}{dx} = \frac{ 3{x}^{2} + 5}{ 3 {y}^{2} - 1 }$

We differentiate again using the quotient rule:

$\frac{d^{2} y}{d {x}^{2} } = \frac{ 6 {x} (3 {y}^{2} - 1) - (3 {x}^{2} + 5)(6y \frac{dy}{dx}) }{ (3 {y}^{2} - 1)^{2} }$

At (1,2), x=1, and y=2

$\frac{d^{2} y}{d {x}^{2} } = \frac{ 6 { \times 1} (3 { \times 2}^{2} - 1) - (3 { \times 1}^{2} + 5)(6 \times 2 \times \frac{8}{11} ) }{ (3 { \times 2}^{2} - 1)^{2} }$

$\frac{d^{2} y}{d {x}^{2} } = - 0.032$