Question

Let \(f(x, y)=3 x^{2} y e^{x-y}\) and \(g(x, y)=\) \(3 x y^{2} e^{y-x}\). Find each of the following. $$ g_{x}(-2,-2) $$

Short answer

The value of \( g_{x}(-2,-2) \) is 0

Step-by-step solution

Understanding Partial Derivative

To find \( g_{x}(-2,-2) \), we need to first understand what it means. This is the partial derivative of function g with respect to x. This process involves treating y as a constant and differentiating normally with respect to x.

Differentiating g(x, y)

The function g(x, y) is a complex function which involves both multiplication and composite functions. The function is \( g(x, y)=3 x y^{2} e^{y-x} \). The rules applied here are the product rule and chain rule which states \( (uv)' = u'v + uv' \) and \( (f(g(x))' = f'(g(x)) * g'(x) \) respectively. Applying these rules, the derivative is calculated as: \( g_{x}=3 y^{2} e^{y-x} - 3 x y^{2} e^{y-x} \). In this step, treat y as a constant. This is a key feature of partial differentiation.

Substituting Values

At this point, we have obtained the function \( g_{x} \) which represents the partial derivative of the function with respect to x. The next step is to substitute the given values of x and y, which are both -2 into the function. \( g_{x}(-2,-2)=3 (-2)^{2} e^{-2-(-2)} - 3 (-2) (-2)^{2} e^{-2-(-2)} \). The result simplifies to 0. Kazaa!.

Key concepts

Functions of Multiple Variables

Functions of multiple variables involve mathematical expressions that depend on two or more independent variables. These functions can be visualized as surfaces if they are defined by two variables, like geographies on maps, with each combination of \(x\) and \(y\) leading to different values of the function output. In our case, \( g(x, y) = 3 x y^{2} e^{y-x} \), the function depends on both \(x\) and \(y\). It showcases how changing the values of two variables can affect the outcome.
Studying such functions allows us to understand how changes in one variable could lead to changes in the function value, by either manipulating one variable while keeping the other constant or altering both. In predictive models or physics, it's essential to grasp how varying these variables interact with each other. This understanding forms the basis for calculating partial derivatives, such as \( g_{x}(-2, -2) \), where we consider changes in the function \(g(x, y)\) dependent primarily on \( x\).

Product Rule

The product rule is a fundamental tool in calculus used for finding the derivative of products of two functions. If we have two functions \(u(x)\) and \(v(x)\) being multiplied, the derivative of their product \(uv\) with respect to \(x\) is given by:

• \[ (uv)' = u'v + uv' \]

In our problem, to take the partial derivative of \( g(x, y) = 3 x y^{2} e^{y-x} \) with respect to \(x\), we treated it as a product of different functions. Here, \(u(x) = 3x\) and \(v(x) = y^{2}e^{(y-x)}\).
First, calculate the derivative of \(u(x)\), which is \(3\), then multiply by \(v(x)\). Next, \(u(x)\) is kept as is, and the derivative of \(v(x)\) is found using chain rule principles. This results in a new expression representing the rate at which the product changes concerning \(x\). Recall, the other variable, \(y\), is held constant.

Chain Rule

The chain rule is another cornerstone of calculus, allowing differentiation of composed functions. It provides a method to handle derivatives when functions are nested within one another. Suppose we have a composition \( f(g(x)) \), the chain rule states that the derivative is given by:

• \[ (f(g(x)))' = f'(g(x)) \cdot g'(x) \]

In the context of our exercise, applying the chain rule helps us differentiate the exponential part of \( g(x, y) = 3 x y^{2} e^{y-x} \). Here, the inner function and the outer function are dictated by the exponential \( e^{(y-x)}\).
To find the derivative with respect to \(x\), we first acknowledge that \( e^{(y-x)}\) depends on \(y-x\). We use the chain rule to find \( e^{(y-x)}' \):

• Outer derivative: \( e^{(y-x)} \rightarrow e^{(y-x)} \)
• Inner derivative: \( -x \rightarrow -1 \)
• Thus, \( e^{(y-x)}' \rightarrow -e^{(y-x)} \)

This step is crucial for adjusting the exponential function’s impact in \(g(x, y)\) when moving slightly in the direction of \(x\), guiding us to accurately express partial derivatives.