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\(43-44\) Find \(\partial z / \partial x\) and \(\partial z / \partial y\) $$ \begin{array}{ll}{\text { (a) } z=f(x) g(y)} & {\text { (b) } z=f(x y)} \\\ {\text { (c) } z=f(x / y)}\end{array} $$

Short Answer

Expert verified
(a) \(\frac{\partial z}{\partial x} = g(y)f'(x), \frac{\partial z}{\partial y} = f(x)g'(y)\); (b) \(\frac{\partial z}{\partial x} = yf'(xy), \frac{\partial z}{\partial y} = xf'(xy)\); (c) \(\frac{\partial z}{\partial x} = \frac{f'(x/y)}{y}, \frac{\partial z}{\partial y} = -\frac{x}{y^2}f'(x/y)\).

Step by step solution

01

Understand the Problem

We need to find the partial derivatives \(\partial z / \partial x\) and \(\partial z / \partial y\) for three different functions: \(z=f(x) g(y)\), \(z=f(x y)\), and \(z=f(x / y)\). Partial derivatives measure how a function changes as one of the variables changes while keeping the other variable constant.
02

Partial Derivative for (a) \(z = f(x)g(y)\)

For \(z = f(x) g(y)\):- To find \(\partial z / \partial x\), treat \(g(y)\) as a constant and differentiate \(f(x)\): \[ \frac{\partial z}{\partial x} = g(y) \cdot f'(x) \]- To find \(\partial z / \partial y\), treat \(f(x)\) as a constant and differentiate \(g(y)\): \[ \frac{\partial z}{\partial y} = f(x) \cdot g'(y) \]
03

Partial Derivative for (b) \(z = f(xy)\)

For \(z = f(xy)\):- To find \(\partial z / \partial x\), use the chain rule. Let \(u = xy\), then \[ \frac{\partial z}{\partial x} = f'(xy) \cdot \frac{\partial}{\partial x}(xy) = y \cdot f'(xy) \]- To find \(\partial z / \partial y\), use the chain rule similarly. \[ \frac{\partial z}{\partial y} = f'(xy) \cdot \frac{\partial}{\partial y}(xy) = x \cdot f'(xy) \]
04

Partial Derivative for (c) \(z = f(x/y)\)

For \(z = f(x/y)\):- To find \(\partial z / \partial x\), use the chain rule. Let \(u = x/y\), then \[ \frac{\partial z}{\partial x} = f'(x/y) \cdot \frac{\partial}{\partial x}(x/y) = \frac{f'(x/y)}{y} \]- To find \(\partial z / \partial y\), apply the quotient rule and chain rule together: \[ \frac{\partial z}{\partial y} = f'(x/y) \cdot \frac{\partial}{\partial y}(x/y) = -\frac{x}{y^2} \cdot f'(x/y) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multivariable Calculus
Multivariable calculus extends the principles of calculus to more than one variable. Instead of dealing with single-variable functions, multivariable calculus handles functions that depend on two or more variables, like the functions described in your exercise: - \(z = f(x)g(y)\), - \(z = f(xy)\), - \(z = f(x/y)\). When working with these types of functions, it is important to study how changes in one variable can affect the function while the other variables are held constant. This study is achieved through the use of partial derivatives.
Partial derivatives provide insights into the sensitivity of a multivariable function to changes in one of its variables. For example,
  • \(\partial z / \partial x\) means you measure the rate of change of \(z\) with respect to \(x\), keeping all other variables constant,
  • while \(\partial z / \partial y\) measures how \(z\) changes with \(y\) unchanged.
Grasping these ideas will empower students to tackle a wide variety of practical problems in physics, engineering, economics, and beyond.
Chain Rule
The chain rule is a fundamental concept in both single-variable and multivariable calculus. It is especially useful when you have composite functions, where one function is nested within another.
Picture a situation where you need to differentiate a function that itself depends on other variables that are also functions of the variables involved. In the context of your exercise, particularly parts (b) and (c), this becomes evident since:
  • For \(z = f(xy)\), the inner composition is the product \(xy\).
  • For \(z = f(x/y)\), the composition is the quotient \(x/y\).
To apply the chain rule, you differentiate the outer function first while keeping the inner function intact, then multiply by the derivative of the inner function.
For instance, while working with \(z = f(xy)\), to find \(\partial z / \partial x\), you'd calculate:\[\frac{\partial z}{\partial x} = f'(xy) \cdot \frac{\partial}{\partial x}(xy) = y \cdot f'(xy)\]This systematic approach simplifies complex derivative calculations, transforming them into sketched steps that are easier to handle.
Quotient Rule
The quotient rule in calculus is a method for finding the derivative of a function that is the division of two other functions. In multivariable calculus, this comes into play when one of the functions is a quotient of quantities involving your variables.
In the exercise, specifically part (c), you encounter \(z = f(x/y)\). Finding the partial derivative of \(z\) with respect to \(y\) requires an application of both the chain rule and the quotient rule:
  • The chain rule tackles the composition aspect of the problem since \(z\) is a function of \(x/y\).
  • The quotient rule helps handle the differentiation of the \(x/y\) term itself.
The quotient rule says that if you have a function \(u/v\), the derivative performed is:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v(u') - u(v')}{v^2}\]For \(z = f(x/y)\), applying this to find \(\partial z / \partial y\) involves these steps:\[\frac{\partial z}{\partial y} = f'(x/y) \cdot \left(-\frac{x}{y^2}\right)\]Each rule has a part to play, breaking down complex derivatives into smaller, more manageable parts to ensure accurate calculations.

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