91Ó°ÊÓ

Understanding the chain rule is crucial when dealing with functions of multiple variables. It's a fundamental tool in calculus that allows us to differentiate composite functions. When we have a function that is the composition of two or more functions, the chain rule helps us to find the derivative of this composite function. For instance, if we have two functions, say \( u(x, y) \) and \( v(u) \), and we want to find the derivative of \( v \) with respect to either \( x \) or \( y \), we would apply the chain rule:
\[ \frac{\partial v}{\partial x} = \frac{\partial v}{\partial u} \cdot \frac{\partial u}{\partial x} \]
Similarly, for differentiating with respect to \( y \), we would use:
\[ \frac{\partial v}{\partial y} = \frac{\partial v}{\partial u} \cdot \frac{\partial u}{\partial y} \]
This tool is especially powerful in multivariable calculus, where functions often depend on more than one variable.

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. Partial derivatives are a key aspect of this field. Unlike the derivative of a function with only one variable, a partial derivative represents the rate of change of a function with respect to one variable, while keeping all others constant.

In the context of our exercise, \( f(x, y) = \arccos(xy) \), there are two variables, \( x \) and \( y \). To understand how \( f \) changes in the direction of each variable, we compute the partial derivatives \( f_x \) and \( f_y \), which inform us how sensitive \( f \) is to variations in \( x \) and \( y \), respectively.

The gradient of a function is a vector that points in the direction of the steepest ascent of the function at a particular point. Each component of the gradient is one of the function's partial derivatives. In multivariable calculus, finding the gradient gives us crucial information about the function's behavior.

For a function \( f(x, y) \), the gradient is given by:
\[ abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) \]
With the function provided in our exercise, if we were to calculate the gradient at the point \((1,1)\), we would use the partial derivatives evaluated at that point. Since both partial derivatives \( f_x \) and \( f_y \) are \( -1 \) when evaluated at \((1,1)\), the gradient would be \( (-1, -1) \), indicating that the function decreases most rapidly if one moves in the negative direction along both \( x \) and \( y \) axes from the point \((1,1)\).