91Ó°ÊÓ

Partial derivatives are like regular derivatives, but they work with functions of more than one variable. They measure how a function changes as you tweak one variable at a time, keeping the others constant.
Imagine you're at a hill, and you want to see the slope by looking north or walking east. That's similar to calculating partial derivatives. They show how steep the function is along different paths.

When dealing with a function of two variables, like our exercise with function \(f(x, y)=x y+\frac{2}{x}+\frac{4}{y}\), finding the partial derivatives \(f_x\) and \(f_y\) helps locate critical points. In our case, \(f_x = y - \frac{2}{x^2}\) indicates how \(f\) changes when \(x\) changes, and \(f_y = x - \frac{4}{y^2}\) shows the change when \(y\) changes.

• Calculate \(f_x\) by differentiating with respect to \(x\).
• Calculate \(f_y\) by differentiating with respect to \(y\).

These calculations are essential for finding critical points.

Critical points in multivariable calculus are the spots where the function might have peaks, valleys, or flat areas. They occur where all first partial derivatives are zero or do not exist.
Think of critical points like crossroads where you decide if you're at a high point, low point, or neither. These points are the candidates for further analysis to find extremas or saddle points.

To find critical points for our function \(f(x, y)\), we solve the system where \(f_x = 0\) and \(f_y = 0\):

• Solve \(y - \frac{2}{x^2} = 0\).
• Solve \(x - \frac{4}{y^2} = 0\).

From this, we find the critical point \((1, 2)\). This point is our candidate for being a relative extrema or saddle point.

The Second Derivative Test helps us discover the nature of the critical points we've found. It decides if they are relative maxima, minima, or saddle points by analyzing the second partial derivatives.
The test works similarly to looking at a hill or a bowl: are you on a peak, trough, or saddle-like area?

For our function, we calculate:

• Second partial derivatives \(f_{xx}, f_{yy},\) and \(f_{xy}\).
• Evaluate these derivatives at the critical point.

Using our values:

• \(f_{xx}(1,2) = 4\), \(f_{yy}(1,2) = 1\), and \(f_{xy}(1,2) = 1\).
• Compute the determinant \(D = f_{xx}f_{yy} - (f_{xy})^2\).

If \(D > 0\), look at \(f_{xx}\): if \(f_{xx} > 0\), it's a minimum; if \(f_{xx} < 0\), it's a maximum.

Relative extrema are the points where a function reaches a local maximum or minimum. They're like the top or bottom of a nearby area, but not necessarily the highest or lowest points overall.
In calculations, they tell us if the critical point we found is higher or lower than directly adjacent points.

In our exercise, once the Second Derivative Test is performed and the determinant \(D\) is calculated, we notice:

• \(D = 3\), which is greater than zero.
• \(f_{xx}(1,2) = 4\), which is positive.

This information confirms that the critical point at \((1, 2)\) is a relative minimum. The landscape around this point forms a bowl-like shape, indicating a low area relative to the surrounding region.