Problem 45 - Calculus Early Transcendentals

Question

Find the indicated derivative for the following functions. \(\frac{\partial z}{\partial x},\) where \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Step-by-Step Solution

Answer

Question: Find the partial derivative of \(z\) with respect to \(x\) for the given equation: \((x\cdot y\cdot z)+ (x\cdot z\cdot y) + (y\cdot x\cdot z) = x\cdot y\cdot z\). Answer: \(\frac{\partial z}{\partial x}= \frac{y}{(1+xy)^2}\)

Step 1: Solve for z

To find the expression for \(z\) in terms of \(x\) and \(y\), we can first rewrite the equation as \((x\cdot y\cdot z)+ (x\cdot z\cdot y) + (y\cdot x\cdot z) = x\cdot y\cdot z\). Then proceed to solve: \(\frac{xyz}{x}+\frac{xyz}{y}+\frac{xyz}{z}=xyz\) \(yz+xz+xy=xyz\) \(z(1+xy)=xyz\) Now, divide by \((1 + xy)\): \(z=\frac{xyz}{1 + xy}\)

Step 2: Calculate \(\frac{\partial z}{\partial x}\)

Now, we are ready to find the partial derivative of \(z\) with respect to \(x\). We will use the quotient rule, which states that if \(u(x)\) and \(v(x)\) are functions of \(x\) then: \(\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\) Here, \(u(x) = xy\), \(v(x)=1+xy\), \(u'(x) = y\) (y is constant with respect to x) and \(v'(x) = y\). Now, apply the quotient rule: \(\frac{\partial z}{\partial x} = \frac{y(1+xy)-xy\cdot y}{(1+xy)^2}\)

Step 3: Simplify the expression

Finally, we need to simplify the expression: \(\frac{\partial z}{\partial x} = \frac{y+xy^2-xy^2}{(1+xy)^2}\) And after simplifying, we get: \(\frac{\partial z}{\partial x} = \frac{y}{(1+xy)^2}\) So the indicated derivative is: \(\frac{\partial z}{\partial x}= \frac{y}{(1+xy)^2}\)

Key Concepts

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. It is particularly useful when dealing with problems involving division, as it allows us to differentiate the numerator and denominator separately. Understanding the Quotient Rule requires knowing that if we have a function represented as \( \frac{u(x)}{v(x)} \), then the derivative of this function is calculated as follows:

\[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]This formula tells us how to handle both parts of the fraction carefully:

• \(u'(x)\): The derivative of the numerator.
• \(v'(x)\): The derivative of the denominator.
• The subtraction \(u'(x)v(x) - u(x)v'(x)\) gives the combined effect of changes in both the numerator and denominator.
• The denominator \((v(x))^2\) is squared to ensure proper behavior of the function as it changes.

In the given exercise, the functions were \(u(x) = xy\) and \(v(x) = 1 + xy\), with their respective derivatives \(y\) and \(y\), which we plugged into the formula to find \(\frac{\partial z}{\partial x}\). By applying these steps, we ensured that the changes in both parts of the quotient were properly accounted for.

Implicit Differentiation

Implicit Differentiation is a method used when it’s difficult or impossible to solve for one variable explicitly before differentiating. Instead of isolating the variable, we differentiate the expression as it is, accounting for all the variables involved. This comes in handy when dealing with equations that represent relationships rather than explicit functions. Here's what you do:

• Differentiate each term with respect to the variable of interest.
• Apply the chain rule as necessary, especially if the function has mixed terms involving both dependent and independent variables.
• Solve for the derivative of the desired variable.

In our exercise, the original equation \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\) did not express \(z\) explicitly in terms of \(x\) and \(y\). By utilizing implicit differentiation, we were able to differentiate directly with respect to \(x\) while treating \(y\) as a constant, eventually leading us to use the Quotient Rule for the simplification.

Multivariable Calculus

Multivariable Calculus deals with functions that depend on more than one variable. It extends the concepts of single-variable calculus to higher dimensions, allowing us to handle functions like \(z = f(x, y)\). A critical understanding is necessary to deal with:

• Partial Derivatives: These are derivatives taken with respect to one variable while keeping the others constant, denoted by symbols like \(\frac{\partial}{\partial x}\).
• Chain Rule in Several Variables: This becomes more complex as it involves derivatives in multiple directions.
• Concepts like tangent planes and gradients, which provide geometrical insights into the behavior of multivariable functions.

In this exercise, the function \(z\) depended on both \(x\) and \(y\), and we were tasked with finding \(\frac{\partial z}{\partial x}\), showing the role of partial derivatives in understanding the change in \(z\) as \(x\) varied. This concept is foundational in understanding how surfaces behave in multidimensional spaces.