91Ó°ÊÓ

Partial derivatives help us understand how a multivariable function changes as we vary one variable while keeping others constant. In the context of our function, which is of the form \(h(x, y) = f(x)g(y)\), we want to understand how it behaves with small changes in \(x\) and \(y\).

To find the partial derivative \(\frac{\partial h}{\partial x}\), treat \(y\) as a constant. The derivative of \(f(x)g(y)\) with respect to \(x\) is \(g(y)f'(x)\). Similarly, for \(\frac{\partial h}{\partial y}\), treating \(x\) as a constant, we get \(f(x)g'(y)\).

At the point \((0,0)\), these derivatives become \(g(0)f'(0)\) for \(\frac{\partial h}{\partial x}\) and \(f(0)g'(0)\) for \(\frac{\partial h}{\partial y}\). Given that \(f(0) = 0\) and \(g(0) = 0\), both derivatives are zero, which indicates a critical point.

The Hessian matrix is a square matrix that organizes all the second-order partial derivatives of a function. In essence, it's a compact representation of how the function curves in different directions.

For our function \(h(x, y)\), the Hessian matrix at any point is:

• \(\frac{\partial^2 h}{\partial x^2} = g(y)f''(x)\)
• \(\frac{\partial^2 h}{\partial y^2} = f(x)g''(y)\)
• \(\frac{\partial^2 h}{\partial x \partial y} = f'(x)g'(y)\)

The matrix is given by:\[\begin{bmatrix}\frac{\partial^2 h}{\partial x^2} & \frac{\partial^2 h}{\partial x \partial y} \\frac{\partial^2 h}{\partial x \partial y} & \frac{\partial^2 h}{\partial y^2}\end{bmatrix}\]

Calculating this at the point \((0,0)\), where both \(f(0)\) and \(g(0)\) equal zero, simplifies the diagonal terms to zero, resulting in the matrix:\[\begin{bmatrix}0 & f'(0)g'(0) \f'(0)g'(0) & 0\end{bmatrix}\]

The determinant of the Hessian matrix can tell us about the curvature of the function at a point. For the 2x2 Hessian matrix:\[\text{Determinant} = \begin{vmatrix}0 & f'(0)g'(0) \f'(0)g'(0) & 0\end{vmatrix}\]

This evaluates as \(0 \times 0 - (f'(0)g'(0))^2 = -(f'(0)g'(0))^2\). Since both \(f'(0)\) and \(g'(0)\) are non-zero and products of these derivatives squared are positive, we conclude that \(-(f'(0)g'(0))^2 < 0\).

A negative determinant suggests that the graph of \(h(x,y)\) near \((0,0)\) has a saddle-like shape, crossing zero in such a way that one direction curves upwards (like a valley) and the other downwards (like a hill).

A saddle point is a critical point where the function doesn't just have a local minimum or maximum, but rather a combination, similar to a horse saddle. In our case study, we apply this concept to \(h(x, y) = f(x)g(y)\).

From the previous sections, we know:

• The first partial derivatives \(abla h(0,0) = (0,0)\), indicating a critical point.
• The Hessian determinant is negative, confirming curvature change through the point.

These factors solidify \((0,0)\) as a saddle point. At this point, neither a pure peak nor a pure valley exists.

This is because the function behaves differently along different slices or cross-sections through \((0,0)\). This mixed behavior characterizes a saddle point, often visible in surfaces resembling a saddle's shape in geometry.