91Ó°ÊÓ

Multivariable calculus expands the field of calculus to functions of several variables. Unlike single-variable calculus where the functions depend on one variable, multivariable calculus deals with functions like \( f(x, y) \) that depend on more than one variable—here, on both \( x \) and \( y \). Calculating partial derivatives is a key task in multivariable calculus, where we differentiate a function with respect to one variable while keeping other variables constant.

As it can be seen in the exercise for \( f(x, y) = 3 \sin(2x + y) - 4 \cos(x - y) \), we first find the first-order partial derivatives with respect to \( x \) and \( y \), denoted \( f_x \) and \( f_y \), treating other variables as constants. These first-order derivatives lead us towards understanding the gradient of the function, or the direction and rate at which the function increases most rapidly.

The chain rule is a fundamental rule in differentiation, including multivariable calculus, that enables us to calculate the derivative of composite functions. In the context of multivariable calculus, we use the chain rule to differentiate functions of functions.

For example, to find \( f_x \) in our exercise, we differentiated the sine and cosine functions that contain expressions of both \( x \) and \( y \) by treating \( y \) as a constant. The chain rule was essential here as the argument of these trigonometric functions also needed to be differentiated with respect to \( x \). The correct application of the chain rule ensures that all nuances of differentiation are respected when dealing with such composite functions.

Once the first-order partial derivatives are calculated, one can proceed to find second-order partial derivatives. These describe the curvature of the function and how the slope changes as we move along a particular direction in the graph of the function. For functions of two variables \( f(x, y) \), we can have three second-order partial derivatives: \( f_{xx} \), \( f_{yy} \) and the mixed derivatives \( f_{xy} \) and \( f_{yx} \).

In our exercise, after finding \( f_x \) and \( f_y \), we differentiated them again with respect to \( x \) and \( y \) respectively to find \( f_{xx} \) and \( f_{yy}\). Mixed derivatives like \( f_{xy} \) and \( f_{yx} \) are obtained by switching the order of differentiation. One of Clairaut's theorem stipulations is that, for smooth functions, the mixed partial derivatives are equal, that is \( f_{xy} = f_{yx} \) which happens to be true for our example.

Differentiation is the process of finding the derivative of a function, which measures how a function changes as its input changes. It’s a fundamental concept in calculus that is extended in multivariable calculus as partial derivatives. These partial derivatives represent the rate of change of the function with respect to one of its variables, holding the others fixed.

In the context of the given exercise, differentiation helps us understand how \( f(x, y) \) changes in the direction of either \( x \) or \( y \). This concept explains why, after finding the first-order derivatives \( f_x \) and \( f_y \) by treating the other variable as a constant, we differentiate again to find the second-order derivatives, delving deeper into the behavior of the function in the space defined by \( x \) and \( y \).