find y'' by implicit differentiation.
2x^3 + 3y^3 = 8
The question is asking for the second derivative by implicit diff. That's where I got confused.
thank you so much!!!
To find the second derivative, y'', by implicit differentiation you need to differentiate both sides of the equation with respect to x twice. Here's a step-by-step guide on how to solve this problem:
Step 1: Differentiate the equation with respect to x.
- Differentiate each term of the equation separately.
- The derivative of 2x^3 with respect to x is 6x^2.
- For the term 3y^3, we need to use the chain rule.
- The derivative of y^3 with respect to y is 3y^2.
- Multiply by the derivative of y with respect to x, which is dy/dx.
- The derivative of 8 with respect to x is 0 because it's a constant.
The differentiated equation is: 6x^2 + 3y^2 * dy/dx = 0
Step 2: Differentiate the equation obtained in Step 1 with respect to x again.
- Differentiate each term of the equation separately.
- The derivative of 6x^2 with respect to x is 12x.
- For the term 3y^2 * dy/dx, we need to use the product rule.
- The derivative of 3y^2 with respect to y is 6y.
- Multiply by the derivative of dy/dx with respect to x, which is d^2y/dx^2.
- Multiply by dy/dx.
- The derivative of 0 with respect to x is 0.
The differentiated equation is: 12x + 6y * dy/dx + 3y^2 * d^2y/dx^2 = 0
Step 3: Solve for d^2y/dx^2.
- Rearrange the equation to isolate d^2y/dx^2.
- Subtract 12x from both sides of the equation.
- Divide both sides of the equation by 3y^2.
The final result is: d^2y/dx^2 = (-12x / 3y^2) - (6y * dy/dx) / 3y^2
So, the second derivative, y'', of the given equation 2x^3 + 3y^3 = 8, is given by (-12x / 3y^2) - (6y * dy/dx) / 3y^2.
6x^2 + 9y^2 y' = 0
y' = -2x^2 / 3y^2