Problem 8 - APEX Calculus

Question

In Exercises \(5-8,\) find the total differential \(d z\). $$ z=x e^{x+y} $$

Step-by-Step Solution

Answer

The total differential is \(dz = e^{x+y}(1 + x)dx + xe^{x+y}dy\).

Step 1: Understand the Problem

We are asked to find the total differential, \(dz\), of the function \(z = xe^{x + y}\). This involves determining how small changes in \(x\) and \(y\) affect the function \(z\).

Step 2: Recall the Formula for Total Differential

The total differential \(dz\) of a function \(z = f(x, y)\) is given by \(dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy\). To find \(dz\), calculate the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

Step 3: Calculate Partial Derivative with Respect to x

Differentiate \(z = xe^{x+y}\) with respect to \(x\) while keeping \(y\) constant. Using the product rule, we get: \(\frac{\partial z}{\partial x} = e^{x+y} + xe^{x+y} = e^{x+y}(1 + x)\).

Step 4: Calculate Partial Derivative with Respect to y

Differentiate \(z = xe^{x+y}\) with respect to \(y\) while keeping \(x\) constant. The derivative is: \(\frac{\partial z}{\partial y} = xe^{x+y}\), because \(x\) is a constant and \(e^{x+y}\) differentiates to itself.

Step 5: Combine Partial Derivatives to Find Total Differential

Substitute the partial derivatives into the total differential formula: \(dz = (e^{x+y}(1 + x))dx + (xe^{x+y})dy\).

Step 6: Present the Final Expression

The total differential of the function \(z = xe^{x+y}\) is given by \(dz = e^{x+y}(1 + x)dx + xe^{x+y}dy\).

Key Concepts

Partial Derivatives

In multivariable calculus, the concept of partial derivatives is crucial. It involves finding the derivative of a function with respect to one variable while keeping other variables constant. This is unlike single-variable calculus, where changes are considered with respect to only one independent variable.
For a function like \(z = xe^{x+y}\), partial derivatives allow us to understand how the function changes as each independent variable (in this case, \(x\) and \(y\)) changes, individually. The notation \(\frac{\partial z}{\partial x}\) signifies the partial derivative of \(z\) with respect to \(x\).

Partial derivatives shed light on how \(z\) behaves:

• When you take the partial derivative of \(z\) with respect to \(x\), it tells you how \(z\) changes as \(x\) changes, holding \(y\) constant.
• Similarly, the partial derivative with respect to \(y\), \(\frac{\partial z}{\partial y}\), shows how \(z\) changes with \(y\), holding \(x\) constant.

Product Rule

The product rule is a fundamental technique in calculus used when differentiating functions that are products of two simpler functions. When dealing with partial derivatives, the product rule becomes essential, as seen in our example function: \(z = xe^{x+y}\).

To apply the product rule, consider \(z = u\times v\), where \(u = x\) and \(v = e^{x+y}\). The product rule states:

• The derivative of \(uv\) with respect to \(x\) is \(u\frac{\partial v}{\partial x} + v\frac{\partial u}{\partial x}\).

Using this principle,

• \(\frac{\partial z}{\partial x} = e^{x+y} + xe^{x+y}\).
• First, differentiate \(e^{x+y}\) with respect to \(x\), treating \(y\) as a constant, which yields \(e^{x+y}\).
• Then add the product of \(x\) and the same derivative, \(e^{x+y}\), showcasing the actual application of the rule.

This results in: \(\frac{\partial z}{\partial x} = e^{x+y}(1+x)\). This illustrates how the product rule helps in breaking down and differentiating complex functions.

Multivariable Calculus

Multivariable calculus is an extension of the basic calculus principles into functions with multiple variables. Unlike single-variable calculus, which deals with one independent variable, multivariable calculus considers scenarios where functions depend on two or more variables.

The function \(z = xe^{x+y}\) operates in a world of multivariable calculus, making concepts like partial derivatives and total differentials extremely relevant. Here's why multivariable calculus matters:

• It allows us to explore how functions change with individual variables in complex spaces.
• Understanding these changes require tools like partial derivatives, which we covered before.

Calculating the total differential \(dz\) in the exercise is a step further into multivariable calculus:

• It combines the individual effects of changes in each variable (\(x\) and \(y\) in this case) on the function \(z\).
• The formula for the total differential \(dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy\) encapsulates how changes in multiple directions combine to affect the function's overall change.

Multivariable calculus offers these unique insights, vital for more advanced applications in fields like physics, engineering, and economics.