91Ó°ÊÓ

Partial derivatives are like stepping stones in multivariable calculus. They help us understand how a function changes with respect to one variable while keeping the other variables constant. In this problem, our goal is to find the partial derivatives of the function \( f(x, y) = e^{2x} \cos y \). The two variables we are focusing on are \( x \) and \( y \). So, we'll look at how changes in only \( x \) affect the function when \( y \) is constant, and vice versa. This is done by taking the partial derivative with respect to \( x \) and \( y \).

• Partial with respect to \( x \): Imagine \( y \) is just a number. Differentiating \( e^{2x} \cos y \) gives us \( f_x = 2e^{2x} \cos y \).
• Partial with respect to \( y \): Now think of \( x \) as a fixed number. Differentiating \( e^{2x} \cos y \) leads to \( f_y = -e^{2x} \sin y \).

These partial derivatives provide the rate of change of the function in the directions of each variable. They are crucial in finding critical points, as we'll see next.

Critical points are places where something interesting happens to the function, such as a peak or a trough. However, they can also be places where the function levels out or forms a saddle shape. To find these points, set each partial derivative to zero.

• For \( f_x = 0 \): \( 2e^{2x} \cos y = 0 \). Since \( e^{2x} > 0 \) for all \( x \), we set \( \cos y = 0 \). This gives \( y = \frac{\pi}{2} + n\pi \), where \( n \) is an integer.
• For \( f_y = 0 \): \(-e^{2x} \sin y = 0 \). Similarly, \( e^{2x} > 0 \) always, so \( \sin y = 0 \). This results in \( y = m\pi \), where \( m \) is an integer.

The critical points occur where these two conditions overlap. By comparing solutions, we conclude that the common points are \( y = \frac{\pi}{2} + (2k+1)\pi \), with \( k \) being an integer. These critical points could be peaks, troughs, or saddle points depending on further tests.

The Second Derivative Test is like a detective that helps us figure out the true nature of critical points. Are they peaks, troughs, or saddle points? To proceed, we calculate the second-order partial derivatives of our function \( f(x, y) = e^{2x} \cos y \).

• \( f_{xx} \): \( 4e^{2x} \cos y \)
• \( f_{yy} \): \( -e^{2x} \cos y \)
• \( f_{xy} \): \( -2e^{2x} \sin y \)

These second derivatives are used to build the Hessian matrix. The determinant of this matrix, given by \( D = f_{xx}f_{yy} - (f_{xy})^2 \), helps classify the critical points. For our example, we find:
\[ D = -4e^{4x} \]This value is less than zero, which indicates that the critical points are saddle points. They aren't peaks or troughs, but rather points where the function curves upward in one direction and downward in another. This is what makes saddle points so fascinating and unique.