Problem 16 - 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. $$ f_{y}(x, y) $$

Step-by-Step Solution

Answer

The partial derivative of the function f with respect to y is \(f_y(x, y) = 3x^2e^{x-y}(1 - y)\)

Step 1: Differentiation of \(f(x, y)\) with respect to y

Using the rule of derivatives, we notice that the function \(f(x, y) = 3x^2 . y . e^{x-y}\) is a product of three terms: \(3x^2\), \(y\) and \(e^{x-y}\). To find the derivative of this function with respect to y, we will need the product rule of differentiation, which states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Applying this rule for three functions we get: \(f_y(x, y) = 3x^2 . 1 . e^{x-y} + 3x^2 . y . e^{x-y} . 0 + 3x^2 . y . e^{x-y} . -1\)

Step 2: Simplifying the expression

The expression simplifies as: \(f_y(x, y) = 3x^2 . e^{x-y} - 3x^2 . y . e^{x-y}\). Keeping constant terms together, it further simplifies to: \(f_y(x, y) = 3x^2e^{x-y}(1 - y)\).

Key Concepts

Multivariable Calculus

Multivariable calculus involves extending calculus concepts to functions of several variables. This area of mathematics is crucial for understanding how variables interact in complex systems. When dealing with functions like \( f(x, y) = 3x^2y e^{x-y} \), we need to compute derivatives with respect to multiple variables. These derivatives, known as partial derivatives, help us explore how a single variable influences the function while keeping others constant.
In the context of this exercise, we are focusing on the partial derivative of \( f \) with respect to \( y \). This approach requires treating \( x \) as a constant while differentiating \( f \) with respect to \( y \). Understanding partial derivatives is essential for analyzing surfaces and curves in higher dimensions. They offer insights into the behavior of multivariable functions at specific points.

Product Rule

The product rule is a fundamental differentiation rule in calculus, especially valuable for dealing with the multiplication of functions. It states that for two functions \( u(x) \) and \( v(x) \), their derivative is given by \( (u \, v)' = u' \, v + u \, v' \). However, in multivariable contexts, functions often involve more than two factors.
In this exercise, the function \( f(x, y) = 3x^2y e^{x-y} \) is a product of three separate components. Therefore, when applying the product rule to compute the partial derivative \( f_y(x, y) \), we consider:

• The derivative of \( y \) with respect to itself, which is 1.
• The constancy of \( 3x^2 \) when differentiating with respect to \( y \).
• The derivative of the exponential component \( e^{x-y} \) which results in \( -e^{x-y} \) as per the chain rule.

Each term is handled carefully to ensure precision, culminating in the expression simplification.

Exponential Functions

Exponential functions, characterized by the form \( e^x \), are unique because they are their own derivatives. This property makes them both fascinating and challenging in calculus.
In the given function \( f(x, y) = 3x^2y e^{x-y} \), the term \( e^{x-y} \) introduces complex differentiation characteristics. When taking the derivative concerning \( y \), the expression \( e^{x-y} \) contributes a factor of \( -1 \), following the chain rule by differentiating the exponent \( (x-y) \) with respect to \( y \).
Understanding how exponential functions behave in differentiation, especially in multiple variable contexts, helps in crafting a clear simplification of expressions, unveiling how these functions stretch or compress with variable changes.