Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, find all first partial derivatives of each function. \(f(x, y)=y \cos \left(x^{2}+y^{2}\right)\)

Step-by-Step Solution

Answer

The partial derivatives are: \( \frac{\partial f}{\partial x} = -2xy \sin(x^2 + y^2) \) and \( \frac{\partial f}{\partial y} = \cos(x^2 + y^2) - 2y^2 \sin(x^2 + y^2) \).

Step 1: Identify the Variables

The function is given by \( f(x, y) = y \cos(x^2 + y^2) \). It is a function of two variables: \( x \) and \( y \). We need to find the first partial derivatives with respect to each variable.

Step 2: Find Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. The function can be seen as \( y \cdot \cos(u) \) where \( u = x^2 + y^2 \). Use the chain rule: \[ \frac{\partial }{\partial x} (y \cdot \cos(u)) = y \cdot \frac{d}{du} (\cos(u)) \cdot \frac{\partial u}{\partial x} = -y \sin(u) \cdot 2x \]. So, \( \frac{\partial f}{\partial x} = -2xy \sin(x^2 + y^2) \).

Step 3: Calculate Partial Derivative with Respect to y

To determine \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant. Apply the product rule: \( \frac{\partial }{\partial y} \left( y \cdot \cos(x^2 + y^2) \right) = \frac{\partial y}{\partial y} \cdot \cos(x^2 + y^2) + y \cdot \frac{\partial }{\partial y} \cos(x^2 + y^2) \). The first term is \( \cos(x^2 + y^2) \). The second term is \( y \cdot (-\sin(x^2 + y^2)) \cdot 2y \) which simplifies to \( -2y^2 \sin(x^2 + y^2) \). Combine: \( \frac{\partial f}{\partial y} = \cos(x^2 + y^2) - 2y^2 \sin(x^2 + y^2) \).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus when dealing with composite functions. A composite function is essentially a function within another function. For example, let's consider a function like our example:

• In the function \( f(x, y) = y \cos(x^2 + y^2) \), the inner function is \( u = x^2 + y^2 \).
• The outer function is \( \cos(u) \).

To find the derivative using the chain rule, you compute the derivative of the outer function with respect to the inner function, then multiply it by the derivative of the inner function with respect to the variable of differentiation.
In our case, we differentiate \( \cos(u) \) to get \( -\sin(u) \), then differentiate the inner function \( u \) to get partial derivatives. For a partial derivative with respect to \( x \), this is \( 2x \), and for \( y \), it's \( 2y \).
This is combined to yield the combined effect of each variable's influence on the function's rate of change.

Product Rule

The product rule is useful when differentiating functions that are products of two other functions. In multivariable calculus, the product rule lets us smoothly handle functions like our example using both chain and product rules together.

• Consider the function \( f(x, y) = y \cdot \cos(x^2 + y^2) \).
• Here, we need to view it as the product of \( y \) and \( \cos(x^2 + y^2) \).

The product rule tells us to differentiate one function while keeping the other constant, and then swap, finally summing the results. So:

• First, keep \( \cos(x^2 + y^2) \) constant and differentiate \( y \) simply to \( 1 \), resulting in \( \cos(x^2 + y^2) \).
• Next, keep \( y \) constant, differentiate \( \cos(x^2 + y^2) \) using the chain rule, and multiply the result by \( y \).

When combined for partial derivatives, this gives us the complete picture of how \( y \cdot \cos(x^2 + y^2) \) changes as \( y \) changes.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to functions of several variables. Such functions depend on more than one input, and derivatives with respect to these inputs reveal how the function evolves. For a function of two variables like \( f(x, y) \),

• We consider inputs \( x \) and \( y \) both independently and jointly.
• Partial derivatives allow us to explore how each variable individually influences the function while treating the others as constant.

This approach is crucial when examining how multivariable systems behave.
In practice, this calculus finds use in physical sciences to model scenarios where several factors drive outcomes simultaneously. Differentiating function \( f(x, y) \) by each variable separately gives us insights into these complex interactions.
Hence, understanding multivariable calculus empowers us with tools to analyze and predict changes in real-world systems with numerous variables.

Function of Two Variables

A function of two variables is simply a function that takes two inputs and assigns them to a single output. Think of it like a surface in three-dimensional space. It’s an important construct in both mathematics and the real world, where many phenomena are naturally influenced by multiple inputs.
When dealing with a function like \( f(x, y) = y \cos(x^2 + y^2) \):

• The inputs \( x \) and \( y \) vary independently, yet their combined cooperation determines the outcome.
• Graphically, the function corresponds to a surface, which undulates based on the combined characteristics of \( x \) and \( y \).

Understanding these functions rely on techniques like partial derivatives, allowing us to examine how changes in one variable impact the function while keeping the other constant.
This analysis is essential for mathematical modeling in fields ranging from physics to economics, where it's key to scrutinize how variables contribute to an overall effect.