91Ó°ÊÓ

To start with, partial derivatives are a fundamental concept in calculus. They involve taking the derivative of a function with more than one variable while keeping the other variables constant. In our exercise, the function is \( f(x, y) = \ln(x^2 + y^2) \), which depends on both \( x \) and \( y \). This means that we can differentiate it partially with respect to each of these variables.

When we compute the partial derivative with respect to \( x \), we treat \( y \) as a constant. Following this approach using the chain rule, the partial derivative of \( f \) with respect to \( x \) is:

• \( \frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2} \)

Similarly, to find the partial derivative with respect to \( y \) while keeping \( x \) constant, we have:

• \( \frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2} \)

These derivatives help us understand how the function \( f(x, y) \) changes as we vary one of its variables slightly while keeping the other fixed.

An electrostatic field is a vector field surrounding charged particles. This field indicates the force that a charged particle would experience at each point in space.

In our exercise, the function \( f(x, y) = \ln(x^2 + y^2) \) describes a potential field, and its gradient \( abla f \) represents the electrostatic field. The gradient of \( f \) is the collection of partial derivatives, forming a vector that points in the direction of the steepest increase of the function.

The gradient vector at a point tells us two things:

• The direction that will increase the function's value most rapidly.
• The magnitude of the gradient vector indicates how steep the increase is.

In electrostatics, the field lines are oriented along the gradient. If the function is a potential related to an electrostatic field, the gradient shows the direction of force on a charged particle.

The chain rule is a powerful tool in calculus used for differentiating composite functions. It allows us to differentiate a function based on the differentiation of an inside function and an outside function.

In our problem, to differentiate \( f(x, y) = \ln(x^2 + y^2) \) partially with respect to \( x \), the chain rule facilitates the simplification. Here’s how it works:

• First, recognize \( x^2 + y^2 \) as the inside function.
• Derive the natural logarithm (\( \ln \)) as the outside function.

By applying the chain rule:

• Differentiation of \( \ln(u) \) with respect to \( u \) gives \( \frac{1}{u} \).
• Then multiply this result by the derivative of the inside function \( 2x \), giving us \( \frac{2x}{x^2 + y^2} \).

The chain rule allows these steps to combine seamlessly, providing an efficient method to find partial derivatives in more complex functions.