Problem 16 - College Algebra and 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

\[f_{y}(x, y) = 3 x^{2}e^{x-y} - 3 x^{2} ye^{x-y}\]

Step 1: Understand function \(f(x, y)\)

The function \(f(x, y)=3 x^{2} y e^{x-y}\) is a composition of several functions. It consists of a product of three sub-functions: \(3 x^{2}\), \(y\), and \(e^{x-y}\).

Step 2: Apply the derivative rules

Differentiating term-by-term using the chain rule of derivatives and the product rule, we get: \n\\[f_{y}(x, y) = \frac{∂}{∂y} (3 x^{2} y e^{x-y})\] \n\Applying product rule (uv)' = u'v + uv', where u = \(3 x^{2} y\), v = \(e^{x-y}\), we get: \n\\[f_{y}(x, y) = (3 x^{2} e^{x-y}) + (-3 x^{2} y e^{x-y})\]

Step 3: Simplify the expression

Finally, after simplifying the expression, \n\\[f_{y}(x, y) = 3 x^{2}e^{x-y} - 3 x^{2} ye^{x-y} \] \n\This is the derivative of \(f(x, y)\) with respect to \(y\).

Key Concepts

Chain Rule in Differentiation

Understanding the chain rule is essential for taking derivatives in calculus, especially when dealing with composite functions. A composite function is like a combination lock where turning one dial affects the next. In calculus language, it's when one function is applied to the result of another function.

Suppose you have a function that depends on another function, such as \( h(x) = f(g(x)) \). To take the derivative of \( h(x) \) with respect to \( x \), the chain rule comes into play, telling us that \( h'(x) = f'(g(x)) \cdot g'(x) \). It orders us to first take the derivative of the outer function, leaving the inner function alone for the moment (\( f'(g(x))\)), and then multiply it by the derivative of the inner function \( g'(x) \).

Returning to our exercise, the function \( f(x, y) = 3 x^{2} y e^{x-y} \) involves an exponential component where the exponent itself is a function of \( x \) and \( y \). As such, the derivative with respect to \( y \) will involve applying the chain rule to account for this layered structure.

Product Rule in Derivatives

The product rule is a powerful tool for finding the derivative of products of two functions. It tells us that if you have two functions \( u(x) \) and \( v(x) \) multiplied together to form \( u(x)v(x) \), then the derivative of this product isn't just the sum of the individual derivatives. Rather, it is given by \( (uv)' = u'v + uv' \).

In layman’s terms, you take the derivative of the first function and multiply it by the second function left intact, then add the product of the first function intact and the derivative of the second function. This rule is used in our exercise to find the derivative of \( 3 x^{2} y \) times \( e^{x-y} \), whereas the first part \( 3 x^{2} y \) remains unaffected when taking the derivative with respect to \( y \), but the exponential function undergoes differentiation due to its dependence on \( y \).

Multivariable Calculus

Multivariable calculus is the extension of calculus in one variable to functions with multiple variables. It's about exploring the landscape of surfaces and curves where each point can have different values for each variable. Imagine navigating a terrain with hills and valleys, where your position is determined not just by moving forward or backward but also by moving left or right.

In the realm of multivariable functions, concepts like partial derivatives become essential tools. A partial derivative, denoted by \( \frac{\partial}{\partial x} f \) for instance, measures how the function \( f \) changes as only the \( x \) variable is varied while the other variable is held constant. In the context of the exercise, \( f_{y}(x, y) \) represents the partial derivative of \( f \) with respect to \( y \) when \( x \) is constant, showing how \( f \) changes as \( y \) alone changes.

Exponential Functions

Exponential functions are the mathematical superheroes of growth and decay processes. They're the formulas behind things that grow rapidly like populations, or decay like radioactive substances. An exponential function can be written in the form \( f(x) = a^{x} \) where \( a \) is a constant and \( x \) is the exponent.

One unique feature about any exponential function with the base \( e \) (Euler's number, approximately 2.71828) is its self-derivative property. This means that the rate of change of an exponential function \( e^{x} \) is still \( e^{x} \). In our exercise, we deal with an exponential component \( e^{x-y} \) which changes when we take the derivative with respect to \( y \) because \( y \) is in the exponent. Here, knowing how to handle exponential functions through differentiation is key to solving for \( f_{y}(x, y) \) correctly.