Problem 50 - Calculus Early Transcendentals

Question

Find $$\partial z / \partial x \text { and } \partial z / \partial y$$ (a) \(z=f(x) g(y) \quad\) (b) \(z=f(x y)\) (c) \(z=f(x / y)\)

Step-by-Step Solution

Answer

(a) \(\frac{\partial z}{\partial x}=f'(x)g(y); \frac{\partial z}{\partial y}=f(x)g'(y).\) (b) \(\frac{\partial z}{\partial x}=f'(xy)y; \frac{\partial z}{\partial y}=f'(xy)x.\) (c) \(\frac{\partial z}{\partial x}=f'(x/y)\frac{1}{y}; \frac{\partial z}{\partial y}=-f'(x/y)\frac{x}{y^2}.\)

Step 1: Understand the Function for Part (a)

In part (a), we have the composite function \( z = f(x)g(y) \). This is a product of two separate functions, one depending only on \(x\) and the other only on \(y\). We need to find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).

Step 2: Find \(\frac{\partial z}{\partial x}\) for Part (a)

Differentiate \(z = f(x)g(y)\) with respect to \(x\), treating \(g(y)\) as a constant since it does not depend on \(x\). This gives\[\frac{\partial z}{\partial x} = f'(x)g(y).\]

Step 3: Find \(\frac{\partial z}{\partial y}\) for Part (a)

Differentiate \(z = f(x)g(y)\) with respect to \(y\), treating \(f(x)\) as a constant since it does not depend on \(y\). This results in\[\frac{\partial z}{\partial y} = f(x)g'(y).\]

Step 4: Understand the Function for Part (b)

In part (b), \( z = f(xy) \) is a function of the product \(xy\). We need to employ the chain rule to find the partial derivatives.

Step 5: Find \(\frac{\partial z}{\partial x}\) for Part (b)

Using the chain rule, we differentiate \(z = f(xy)\) with respect to \(x\):\[\frac{\partial z}{\partial x} = f'(xy) \cdot y.\]

Step 6: Find \(\frac{\partial z}{\partial y}\) for Part (b)

Again applying the chain rule, differentiate \(z = f(xy)\) with respect to \(y\):\[\frac{\partial z}{\partial y} = f'(xy) \cdot x.\]

Step 7: Understand the Function for Part (c)

In part (c), the function is \( z = f(x/y) \). This is a function of the ratio \(x/y\). We will again use the chain rule for partial differentiation.

Step 8: Find \(\frac{\partial z}{\partial x}\) for Part (c)

Using the chain rule, differentiate \(z = f(x/y)\) with respect to \(x\):\[\frac{\partial z}{\partial x} = f'(x/y) \cdot \frac{1}{y}.\]

Step 9: Find \(\frac{\partial z}{\partial y}\) for Part (c)

Employ the chain rule to differentiate \(z = f(x/y)\) with respect to \(y\):\[\frac{\partial z}{\partial y} = f'(x/y) \cdot \left(-\frac{x}{y^2}\right).\]

Key Concepts

chain rule

The chain rule is a crucial technique in calculus, particularly when dealing with functions that are compositions of other functions. To understand the chain rule, imagine you have a function that is made up by combining two or more functions. When you want to differentiate this composite function, the chain rule helps you find the derivative by breaking down the problem into simpler parts. For example, if you are tasked with finding the partial derivative of a function like \( z = f(xy) \), you can think of it as first differentiating \( f(u) \) where \( u = xy \), and then multiplying by the derivative of \( u \) with respect to \( x \) or \( y \).

• Apply the chain rule by identifying the inner function: for instance, in \( z = f(xy) \), the inner function is \( xy \).
• Differentiating \( f(u) \) with respect to \( u \) gives you \( f'(u) \).
• Find the derivative of the inner function: \( u = xy \) means \( \frac{du}{dx} = y \) and \( \frac{du}{dy} = x \).
• Combine these using the chain rule: for \( \frac{\partial z}{\partial x} \), it is \( f'(xy) \cdot y \), and for \( \frac{\partial z}{\partial y} \), it is \( f'(xy) \cdot x \).

Understanding and applying the chain rule efficiently allows you to tackle complex differentiation problems with ease. It is particularly useful in multivariable calculus where functions depend on several variables.

product rule

In calculus, the product rule is an essential tool when differentiating functions that are products of two or more other functions. Imagine a scenario where you have a function like \( z = f(x)g(y) \). Here, the product rule is necessary to efficiently find its partial derivatives.

The product rule states that the derivative of a product of functions is not just the product of their derivatives. Instead, you differentiate one function while keeping the other function constant and then add the product of the other derivative. Here's how it works in multivariable calculus:

• Identify the two functions being multiplied, like \( f(x) \) and \( g(y) \) in the example \( z = f(x)g(y) \).
• For \( \frac{\partial z}{\partial x} \): treat \( g(y) \) as a constant, then differentiate \( f(x) \) which gives you \( f'(x)g(y) \).
• For \( \frac{\partial z}{\partial y} \): treat \( f(x) \) as constant, and differentiate \( g(y) \), resulting in \( f(x)g'(y) \).

The product rule simplifies the differentiation of product functions, making it manageable to handle complex expressions in multivariable settings

multivariable calculus

Multivariable calculus extends the principles of single-variable calculus to functions that depend on two or more variables. It increases the complexity of calculus but also its application within fields such as engineering and physics.

In multivariable calculus, functions might depend on several variables like \( x, y, \) or \( z \), making the calculation of derivatives more intricate. Partial derivatives are used to understand how a function changes concerning each variable independently.

• A partial derivative with respect to a particular variable treats other variables as constants.
• For example, in \( z = f(x, y) \), \( \frac{\partial z}{\partial x} \) examines how \( z \) changes as \( x \) changes, keeping \( y \) constant.
• Similarly, \( \frac{\partial z}{\partial y} \) shows the effect of changing \( y \) while holding \( x \) constant.

Dealing with multiple variables allows us to describe surfaces or volumes mathematically and is foundational for fields like optimization, where you often need to understand how changes in various factors affect an outcome or objective function. Multivariable calculus is essential for modeling and solving problems in a three-dimensional space.