91Ó°ÊÓ

Partial derivatives are an essential concept in multivariable calculus. When dealing with a function of several variables, like in the given function \( f(x, y) = 2xy - 5x^2 - 2y^2 + 4x + 4y - 4 \), partial derivatives help us understand how the function changes with respect to one variable at a time while keeping the others constant.

To find the partial derivative with respect to \( x \), denoted as \( f_x \), you differentiate the function treating \( y \) as a constant. In the example, this results in \( f_x = 2y - 10x + 4 \). Likewise, for the partial derivative with respect to \( y \), or \( f_y \), treat \( x \) as a constant, yielding \( f_y = 2x - 4y + 4 \).

Understanding partial derivatives is crucial as they are the foundation for finding critical points and analyzing the behavior and properties of multivariable functions.

Critical points in multivariable calculus are where a function's partial derivatives are zero or undefined. These points can indicate potential locations of local maxima, local minima, or saddle points. To locate these points, equate the partial derivatives \( f_x \) and \( f_y \) to zero and solve the resulting system of equations.

In the given exercise, solving \( f_x = 2y - 10x + 4 = 0 \) and \( f_y = 2x - 4y + 4 = 0 \) leads to finding the critical point \( \left(\frac{2}{3}, \frac{4}{3}\right) \). Solving these linear equations may involve substitution or elimination methods.

Identifying critical points is a key step in analyzing a function's behavior across its domain and lays the groundwork for further investigation using tests like the Second Derivative Test.

The Hessian determinant is a critical tool for assessing the nature of critical points in multivariable functions. It involves the second partial derivatives and helps determine whether a critical point is a local maximum, local minimum, or a saddle point.

The Hessian matrix \( H \) for a function \( f(x, y) \) is defined as:

• \( f_{xx} = \frac{\partial^2 f}{\partial x^2} \)
• \( f_{yy} = \frac{\partial^2 f}{\partial y^2} \)
• \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} \)

For the example, this results in \( f_{xx} = -10 \), \( f_{yy} = -4 \), and \( f_{xy} = 2 \).

The Hessian determinant \( H \) is computed as \( H = f_{xx} f_{yy} - (f_{xy})^2 \). Substituting the values gives \( H = 36 \), used for determining the nature of the critical point through the Second Derivative Test.

The Second Derivative Test is an approach to classify critical points found earlier as local maxima, local minima, or saddle points. It relies on the second partial derivatives and their arrangement in the Hessian determinant.

In our instance, we found the critical point \( \left(\frac{2}{3}, \frac{4}{3}\right) \). The steps to apply the Second Derivative Test are:

• Compute the Hessian determinant \( H \).
• Check the sign of \( H \).
• Consider the sign of \( f_{xx} \).

If \( H > 0 \) and \( f_{xx} < 0 \), the test confirms a local maximum at the critical point. This is the case in the provided solution, indicating the presence of a local maximum at \( \left(\frac{2}{3}, \frac{4}{3}\right) \).

This test is a straightforward yet powerful method to determine the behavior of the function at critical points using second derivative information.