91Ó°ÊÓ

The gradient is a crucial concept when working with directional derivatives of multivariable functions. Think of it as a vector that points in the direction of the greatest rate of increase of a function. For a function like \(g(x, y)\), the gradient, often labeled as \(abla g\), consists of the partial derivatives of the function.

Mathematically, it's expressed as:

• \(abla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right)\)

Here, each component of the gradient vector \(abla g\) is a partial derivative with respect to one of the variables \(x\) or \(y\). In practical terms, it tells you how much the function \(g\) changes as you move in the direction of the respective variable.

For the given problem, we computed the gradient of \(g(x, y) = \frac{x-y}{xy+2}\) at point \(P_0 = (1, -1)\), resulting in the vector \((3, 1)\). This means that at point \(P_0\), the function increases most rapidly in the direction of the vector \(3\mathbf{i} + 1\mathbf{j}\).

Partial derivatives are used to find how a multivariable function changes as one of its variables changes, while keeping the other variables constant. They're the building blocks for gradients.

• The partial derivative with respect to \(x\) is denoted as \(\frac{\partial g}{\partial x}\).
• The partial derivative with respect to \(y\) is denoted as \(\frac{\partial g}{\partial y}\).

To calculate partial derivatives for functions with fractions, like the one in the problem, we use the quotient rule. This involves differentiating the numerator and denominator separately. In our problem, we performed these calculations to get:

• \(\frac{\partial g}{\partial x} = \frac{y + 2 - xy + y^2}{(xy+2)^2}\)
• \(\frac{\partial g}{\partial y} = \frac{x(1-y) + xy}{(xy+2)^2}\)

Substituting the point \(P_{0} = (1, -1)\) into these derivatives, we found the specific rate of change at that point. Partial derivatives help us understand how the function behaves when tweaking just one variable at a time.

A unit vector is simply a vector with a magnitude of 1. It's often used to represent direction, without scaling any vector components. When talking about directional derivatives, we need to ensure the direction vector \(\mathbf{u}\) is a unit vector. This ensures the directional derivative precisely measures the rate of change in the specified direction. To turn a vector into a unit vector, we divide each component by the vector's magnitude:

• The original direction vector \(\mathbf{u} = 12\mathbf{i} + 5\mathbf{j}\)
• The magnitude \(\|\mathbf{u}\| = \sqrt{12^2 + 5^2} = 13\)
• Thus, the unit vector \(\mathbf{u}\) becomes \((\frac{12}{13}, \frac{5}{13})\)

Dividing by the magnitude always gives us a vector that points in the same direction but only at a length of 1. This normalization step is crucial because it allows the directional derivative to represent the true rate at which the function changes in that specific direction.