Problem 41 - Calculus Early Transcendentals

Question

Find the indicated partial derivatives. $$f(x, y, z)=\frac{y}{x+y+z} ; \quad f_{y}(2,1,-1)$$

Step-by-Step Solution

Answer

The partial derivative \( f_y(2, 1, -1) \) is \( \frac{1}{4} \).

Step 1: Identify the Function

The given function is \( f(x, y, z) = \frac{y}{x+y+z} \). We need to find the partial derivative \( f_y(x, y, z) \), which represents how the function changes with respect to \( y \) while keeping \( x \) and \( z \) constant.

Step 2: Apply the Quotient Rule

The function \( f(x, y, z) = \frac{y}{x+y+z} \) is a quotient of two functions, where the numerator is \( y \) and the denominator is \( x+y+z \). To find the partial derivative with respect to \( y \), we apply the quotient rule, which states: \( f_y = \frac{(g(x+y+z)\cdot1) - (y\cdot g'(x+y+z))}{(x+y+z)^2} \), where \( g(x+y+z) \) is the numerator (constant with respect to \( y \)).

Step 3: Differentiate the Numerator

Differentiate the numerator \( y \) with respect to \( y \). This gives us 1, since \( \frac{d}{dy} y = 1 \).

Step 4: Differentiate the Denominator

The derivative of the denominator \( x + y + z \) with respect to \( y \) is 1, because \( x \) and \( z \) are treated as constants, hence don't contribute to the derivative.

Step 5: Simplify the Partial Derivative Expression

Using the quotient rule, the partial derivative \( f_y \) can be simplified as follows: \[ f_y = \frac{(x+y+z)(1) - (y)(1)}{(x+y+z)^2} = \frac{x+y+z-y}{(x+y+z)^2} = \frac{x+z}{(x+y+z)^2} \]

Step 6: Substitute Values

Now substitute \( x = 2 \), \( y = 1 \), and \( z = -1 \) into \( f_y \). This gives: \[ f_y(2, 1, -1) = \frac{2 + (-1)}{(2 + 1 - 1)^2} \]

Step 7: Calculate the Result

Simplify the expression \[ \frac{2 - 1}{(2 + 1 - 1)^2} = \frac{1}{2^2} = \frac{1}{4} \] Hence, \( f_y(2, 1, -1) = \frac{1}{4} \).

Key Concepts

Calculus

Calculus is one of the most important branches of mathematics. It is used to understand how things change. In calculus, we work with derivatives and integrals, which are tools for analyzing the behavior of functions.
Derivatives are used to measure the rate at which a quantity changes. When we talk about derivatives, we often refer to the slope or steepness of a curve. In the context of multivariable functions, this becomes slightly more complex.

• In single-variable calculus, we commonly find the derivative of a function with respect to one variable.
• For functions with multiple variables, such as those seen in many real-world applications, we calculate partial derivatives.

Partial derivatives allow us to observe how a function changes in one particular direction while keeping other variables constant. This is crucial in fields such as physics, engineering, and economics, where systems are usually influenced by several variables at once.

Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of a function that is the ratio of two other functions. When dealing with a quotient, like the function given in the exercise, it's important to apply this rule to correctly find derivatives.
If you have a function expressed as a quotient \( f(x) = \frac{u(x)}{v(x)} \), then the derivative \( f'(x) \) is found using:

• \[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]

In the context of partial derivatives, we apply the quotient rule similarly. The difference is that we're differentiating with respect to one variable, treating all others as constants.

• For the function \( f(x, y, z) = \frac{y}{x+y+z} \), \( y \) is the numerator and \( x+y+z \) is the denominator.
• The partial derivative with respect to \( y \) involves treating \( x \) and \( z \) as constants.

Using the quotient rule, where we consider the derivative of \( y \) is \( 1 \) and the derivative of \( x+y+z \) with respect to \( y \) is also \( 1 \), you can systematically solve for the partial derivative.

Multivariable Functions

Multivariable functions are functions that depend on more than one input variable. Unlike single-variable functions that are represented on a two-dimensional plane, multivariable functions exist in higher dimensions.
Understanding these functions is key to grasping many advanced concepts in calculus, as they are used to describe a wide array of phenomena.

• For a function like \( f(x, y, z) \), each of \( x \), \( y \), and \( z \) contributes to the function's output.
• Partial derivatives help describe how the function behaves with changes to one variable at a time.

In calculating partial derivatives for such functions, we use techniques similar to those in single-variable calculus, but we must be mindful that changes occur in a multi-dimensional space.
This learning scales with practical applications in various domains like

• Optimizing processes in engineering
• Modeling financial systems
• Predicting physical phenomena in science

By effectively using partial derivatives, we can navigate the complexity and harness the full potential of multivariable functions.