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Partial derivatives are a fundamental concept in calculus, particularly for functions of more than one variable. They allow us to explore how the function changes as each individual variable changes, holding the others constant.鈦�
For a function like \[ f(x, y) = x^3 + 3xy^2 - 15x + y^3 - 15y \]we find the partial derivatives with respect to \(x\) and \(y\). This requires taking the derivative of the function one variable at a time, treating other variables as constants.

• Partial derivative with respect to \(x\) is: \( f_x(x, y) = 3x^2 + 3y^2 - 15 \)
• Partial derivative with respect to \(y\) is: \( f_y(x, y) = 6xy + 3y^2 - 15 \)

These derivatives help identify critical points by setting them equal to zero, forming a system of equations. Solving this system enables us to pinpoint locations where the function's rate of change is momentarily halted, which could be potential points of maxima, minima, or saddle points.

After finding critical points from partial derivatives, the second derivative test helps us determine the nature of these points 鈥� whether they are maxima, minima, or saddle points. This test looks at the behavior of the function around critical points using second partial derivatives.To perform the second derivative test, we calculate:

• \( f_{xx} = \frac{\partial^2 f}{\partial x^2} \), which is \(6x\) in our example.
• \( f_{yy} = \frac{\partial^2 f}{\partial y^2} \), which simplifies to \(6y\).
• Mixed derivative \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} \), which is \(6y\) here.

The second derivative test involves using these values in the Hessian matrix, particularly focusing on the Hessian determinant. The process assesses concavity or convexity at those critical points, thus helping conclude if they're peaks, valleys, or none (saddle points).

The Hessian determinant plays a crucial role in the second derivative test for determining the nature of critical points. It is derived from the Hessian matrix, which is a square matrix of second-order partial derivatives of the function.For our function:- We construct the Hessian as: \[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 6x & 6y \ 6y & 6y \end{bmatrix}\]- The determinant of this Hessian matrix is: \[H = f_{xx}f_{yy} - (f_{xy})^2 = (6x)(6y) - (6y)^2\]The value of the Hessian determinant at a critical point helps us decide:

• If \(H > 0\) and \(f_{xx} > 0\), it signifies a local minimum.
• If \(H > 0\) and \(f_{xx} < 0\), it indicates a local maximum.
• If \(H < 0\), the point is a saddle point.

A zero Hessian determinant means the test is inconclusive. By evaluating the Hessian determinant at each critical point, we comprehensively determine their roles in the function's topography.