Questions: If f(x, y) = 2x^2y^3 - 3xy^2 + 2x^2 + 3y^2 + 1, then ∂^2f / ∂y^2 equals 18x^2y^2 - 6xy + 6y 12xy^2 12x^2y - 6x + 6 None of the above

Transcript text: If $f(x, y)=2 x^{2} y^{3}-3 x y^{2}+2 x^{2}+3 y^{2}+1$, then $\partial^{2} f / \partial y^{2}$ equals $18 x^{2} y^{2}-6 x y+6 y$ $12 x y^{2}$ $12 x^{2} y-6 x+6$ None of the above

Solution

Solution Steps

To find the second partial derivative of the function $ f(x, y) = 2x^2 y^3 - 3xy^2 + 2x^2 + 3y^2 + 1 $ with respect to $ y $, we first need to compute the first partial derivative with respect to $ y $, and then differentiate that result again with respect to $ y $.

Step 1: Define the Function

Given the function: \[ f(x, y) = 2x^2 y^3 - 3xy^2 + 2x^2 + 3y^2 + 1 \]

Step 2: Compute the First Partial Derivative with Respect to $ y $

The first partial derivative of $ f $ with respect to $ y $ is: \[ \frac{\partial f}{\partial y} = 6x^2 y^2 - 6xy + 6y \]

Step 3: Compute the Second Partial Derivative with Respect to $ y $

The second partial derivative of $ f $ with respect to $ y $ is: \[ \frac{\partial^2 f}{\partial y^2} = 12x^2 y - 6x + 6 \]