"If 2x^2 + y^2 = 17, then evaluate the second derivative of y with respect to x when x = 2 and y = 3. Round your answer to two decimal places. Use the minus sign, -, for negative values."

To find the second derivative of y with respect to x (denoted as d²y/dx² or y''), we first need to implicitly differentiate the given equation with respect to x, since y is a function of x. The given equation is:

2x² + y² = 17

First derivative (implicit differentiation): Differentiating both sides of the equation with respect to x:

d/dx(2x²) + d/dx(y²) = d/dx(17)

4x + 2yy' = 0 (where y' denotes dy/dx)

Now, we need to find y'. To do that, solve for y':

2yy' = -4x y' = -4x / (2y) y' = -2x / y

Second derivative (implicit differentiation): Now we differentiate y' with respect to x to find y'':

d/dx(y') = d/dx(-2x / y)

Using the quotient rule (d/dx(u/v) = (vdu - udv)/v²):

y'' = d/dx(-2x) * 1/y + 2x * d/dx(1/y) y'' = (-2)y⁻¹ + 2x * (-y' * y⁻²) y'' = -2y⁻¹ - 2xy' * y⁻²

Now plug in the values for x and y':

y'' = -2(1/3) - 2(2)(-2(2)/(3)) * (1/3)² y'' = -2/3 + 8/3 * (1/9) y'' = -2/3 + 8/27

To get a common denominator, multiply -2/3 by 9/9:

y'' = (-2/3)*(9/9) + 8/27 y'' = -18/27 + 8/27 y'' = (-18 + 8)/27 y'' = -10/27

Now, we round this result to two decimal places:

y'' ≈ -0.37