Problem 18 - Calculus an Applied Approach

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. $$ g_{y}(x, y) $$

Step-by-Step Solution

Answer

Therefore, the derivative of the function \(g(x, y)\) w.r.t \(y\) is \(3y^2e^{y - x} + 6xy e^{y - x} + 6x y^2e^{y - x}\).

Step 1: Identify the functions

The function \(g(x, y) = 3xy^2 e^{y - x}\) is a product of three functions: \(u = 3x\), \(v = y^2\), and \(w = e^{y - x}\). The product rule for three factors states that \( (uvw)' = u'vw + uv'w + uvw'.\)

Step 2: Calculate the derivatives

Differentiate each: \(u' = 3\) (constant as we are differentiating w.r.t \(y\)), \(v' = 2y\), and \(w' = e^{y - x}(1 - (-1)) = e^{y - x}(2)\) by the chain rule.

Step 3: Chain rule and assembly

Using the identified functions and their derivatives, substitute into the product rule equation: \(g_y(x, y) = u'vw + uv'w + uvw' = 3y^2e^{y - x} + 3x(2y)e^{y - x} + 3x(y^2)(2)e^{y - x} = 3y^2e^{y - x} + 6xy e^{y - x} + 6x y^2e^{y - x}.\)

Key Concepts

The Chain Rule in Partial Differentiation

The chain rule is a fundamental concept in calculus used extensively in partial differentiation, especially in multivariable calculus. When differentiating a composite function, the chain rule enables us to calculate the derivative effectively by taking into account the rates of change relative to each independent variable involved. In simpler terms, the chain rule helps us understand how a function changes as its input variables change, by linking the derivatives of the outer and inner functions.

For example, when faced with a function like \(w = e^{y - x}\), as we saw in the exercise, applying the chain rule is essential. Since \(w\) depends on \(y - x\), which is a function itself, taking the derivative of \(w\) with respect to \(y\) involves treating \(y - x\) as a single function (let's call it \(h\)), which is then differentiated. We can summarize this as \(\frac{dw}{dy} = \frac{dw}{dh} \cdot \frac{dh}{dy}\), where the derivative of \(h\) with respect to \(y\) is 1, and the derivative of \(w\) with respect to \(h\) is \(e^h\).

The Product Rule in Partial Differentiation

The product rule is a cornerstone for handling derivatives of functions that are products of two or more factors. In partial differentiation, when dealing with multivariable functions, applying the product rule allows us to differentiate each part of the function individually while considering the others as constants.

In the given exercise with the function \(g(x, y) = 3x y^2 e^{y - x}\), we noticed that it's the product of three separate functions. When differentiating \(g\) with respect to \(y\), we apply the product rule for multiple functions. The product rule states that the derivative of the products of functions \(u\), \(v\), and \(w\) is \(u'vw + uv'w + uvw'\). This approach simplifies complex derivatives by breaking them into more manageable parts. We differentiate each function independently and then combine them using the product rule to obtain the partial derivative of \(g\) with respect to \(y\).

Understanding Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions with two or more variables. In this broader context, we investigate how these functions change as we adjust multiple inputs simultaneously. The groundwork of multivariable calculus comprises partial derivatives, directional derivatives, and multiple integrals, among others.

Partial derivatives, such as \(g_y(x, y)\) in our exercise, assess the rate of change of a multivariable function with respect to one variable while keeping the others constant. When computing \(g_y(x, y)\), we used the chain rule and product rule in concert, reflecting the complexity and interconnectedness of concepts within multivariable calculus. Grasping these foundational rules simplifies understanding higher-level concepts like gradient vectors and optimization problems in multivariable functions.