Questions: f(x, y) = x^2 e^(-x^2-y^3) 1. fy = 2 x^3 y^2 e^(-x^2-y^3) 2. fy = 2 x^2 y^3 e^(-x^2-y^3) 3. fy = -3 x^2 y^2 e^(-x^2-y^3) 4. fy = -3 x^3 y^2 e^(-x^2-y^3) 5. fy = 3 x^2 y^2 e^(-x^2-y^3) 6. fy = -2 x^2 y^3 e^(-x^2-y^3)

Solution Steps

To find the partial derivative of the function \( f(x, y) = x^2 e^{-x^2 - y^3} \) with respect to \( y \), we need to apply the chain rule. The exponential function's derivative will involve the derivative of the exponent \(-x^2 - y^3\) with respect to \( y \).

We start with the function given by

\[ f(x, y) = x^2 e^{-x^2 - y^3}. \]

To find the partial derivative of \( f \) with respect to \( y \), we apply the chain rule. The derivative of the exponential function involves the derivative of the exponent:

\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( x^2 e^{-x^2 - y^3} \right). \]

Using the chain rule, we find:

\[ \frac{\partial f}{\partial y} = x^2 \cdot e^{-x^2 - y^3} \cdot \frac{\partial}{\partial y}(-x^2 - y^3) = x^2 \cdot e^{-x^2 - y^3} \cdot (-3y^2). \]

Thus, we can simplify this to:

\[ \frac{\partial f}{\partial y} = -3x^2y^2 e^{-x^2 - y^3}. \]

The final expression for the partial derivative \( f_y \) is:

\[ f_y = -3x^2y^2 e^{-x^2 - y^3}. \]

Final Answer