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The concept of a gradient is central in understanding how a multivariable function, like \( f(x, y) = \frac{x-y}{x+y} \), changes at any given point. To find the gradient, we calculate the partial derivatives of the function, which gives us a vector that points in the direction of the greatest rate of increase of the function. In this case, the gradient \( abla f = \left( \frac{2y}{(x+y)^2}, -\frac{2x}{(x+y)^2} \right) \), is composed of two partial derivatives - one with respect to \( x \), and one with respect to \( y \). - The partial derivative with respect to \( x \) highlights how \( f \) changes as \( x \) changes, holding \( y \) constant.- Similarly, the partial derivative with respect to \( y \) indicates how \( f \) changes with respect to \( y \,\). Evaluating the gradient at the point \( \left( -\frac{1}{2}, \frac{3}{2} \right) \), we obtain \( (3, 1) \). This vector reflects the steepest ascent from that point.

The rate of change in a multivariable function in a specific direction is determined by the directional derivative. The directional derivative, denoted \( D_\mathbf{u} f \), measures how much the function changes as you move in the direction of a given vector \( \mathbf{u} \). In simpler terms, it tells us how the function varies in a particular direction.The directional derivative relates to the gradient since the maximum rate of change of the function occurs in the gradient's direction, whereas the minimum (or largest negative) rate occurs in the opposite direction. If you move perpendicularly to the gradient, the rate of change is zero, meaning the function's value doesn't change in that direction. To utilize this concept:- Calculate the gradient of the function.- Find the unit vector in the desired direction.- Use the dot product of these vectors to determine the directional derivative.

Partial derivatives are the building blocks of the gradient and help in understanding functions of several variables. For a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), gives the rate at which \( f \) changes as \( x \) increases, while keeping \( y \) constant. Similarly, \( \frac{\partial f}{\partial y} \) indicates how \( f \) changes as \( y \) increases with \( x \) fixed.In our example, the function \( f(x, y) = \frac{x-y}{x+y} \) has the partial derivatives:- \( \frac{\partial f}{\partial x} = \frac{2y}{(x+y)^2} \)- \( \frac{\partial f}{\partial y} = -\frac{2x}{(x+y)^2} \)These derivatives, when computed at specific points, reveal insights into how the function behaves and changes in those regions. They're critical for analyzing and understanding how the function responds to small changes in \( x \) and \( y \).

A unit vector is a vector with a magnitude of 1, used to specify a direction. When dealing with directional derivatives, using a unit vector \( \mathbf{u} \) ensures that the rate of change is measured accurately without scaling from the vector's magnitude. In this context, calculating the directional derivative \( D_\mathbf{u} f \) involves normalizing your direction vector.To find a unit vector, take any vector \( \mathbf{v} = (v_x, v_y) \) and divide it by its magnitude \( \| \mathbf{v} \| \). For example, if the direction of maximum increase is parallel to the gradient \( (3, 1) \), the unit vector in this direction is calculated as:\[ \mathbf{u} = \frac{1}{\sqrt{3^2 + 1^2}} (3, 1) = \frac{1}{\sqrt{10}}(3, 1) \]Normalizing a vector makes easy comparisons and calculations possible, allowing for determining directional derivatives along any specified path. Knowing how to create a unit vector is key in applying gradients to real-world problems effectively.