Problem 25 - Calculus Early Transcendentals

Question

\(25-30\) Find the differential of the function. $$ z=x^{3} \ln \left(y^{2}\right) $$

Step-by-Step Solution

Answer

\( dz = (3x^2 \ln(y^2)) dx + \left(x^3 \frac{2}{y}\right) dy \).

Step 1: Understanding the Function

The given function is a product of two separate functions: the power function \(x^3\) and the logarithmic function \(\ln(y^2)\). The problem is asking us to find the differential of this function.

Step 2: Applying the Product Rule

To differentiate the function \(z = x^3 \ln(y^2)\), we need to use the product rule. The product rule states that \((uv)' = u'v + uv'\), where \(u = x^3\) and \(v = \ln(y^2)\). Hence, we differentiate each part separately.

Step 3: Differentiating \(x^3\)

Differentiate \(x^3\) with respect to \(x\). The derivative is \(u' = 3x^2\).

Step 4: Differentiating \(\ln(y^2)\)

Differentiate \(\ln(y^2)\) with respect to \(y\). By applying the chain rule, since \(\ln(y^2) = 2\ln(y)\), the derivative is \(v' = \frac{2}{y}\).

Step 5: Writing the Differential

The differential of \(z\), denoted \(dz\), consists of terms from differentiating with respect to both \(x\) and \(y\):\[ dz = \left(\frac{\partial z}{\partial x} \right) dx + \left(\frac{\partial z}{\partial y}\right) dy \].From the product rule, \(\frac{\partial z}{\partial x} = 3x^2 \ln(y^2)\) and \(\frac{\partial z}{\partial y} = x^3 \cdot \frac{2}{y}\).Thus, \[ dz = (3x^2 \ln(y^2)) dx + \left(x^3 \cdot \frac{2}{y}\right) dy \].

Key Concepts

Product Rule

When dealing with the differentiation of a product of two functions, the product rule is an essential tool. It helps us take the derivative of the product by considering each function separately and then combining their derivatives.
For a product of functions, such as in the given problem with the function \( z = x^3 \ln(y^2) \), we consider one part \( u = x^3 \) and the other part \( v = \ln(y^2) \).

• Differentiate each function separately: Calculate the derivative \( u' \) and \( v' \).
• Apply the product rule: \( (uv)' = u'v + uv' \).
• Combine the results: Insert both derivatives into the rule to get the final result.

In this scenario:

• \( u' = 3x^2 \)
• \( v' = \frac{2}{y} \) (found using the chain rule)

This leads to the differential expression which is put together using the contributions of each part of the function.

Chain Rule

The chain rule is a fundamental principle in calculus, that is used when differentiating a composite function. In our given function, \( \ln(y^2) \) involves an "inner function" \( y^2 \) and an "outer function" \( \ln \).
To differentiate \( \ln(y^2) \), we apply the chain rule:

• The derivative of the outer function, \( \ln(u) \), is \( \frac{1}{u} \).
• The derivative of the inner function, \( y^2 \), is \( 2y \).

By combining these:

• Apply the chain rule: \( \frac{d}{dy}(\ln(y^2)) = \frac{1}{y^2} \cdot 2y \).
• This simplifies to \( \frac{2}{y} \).

Using the chain rule allows us to manage more complex functions and ensures proper calculation of the derivative of nested functions.

Partial Derivatives

Partial derivatives are used when you have functions of multiple variables, to understand how changes in one variable affect the function. For the function \( z = x^3 \ln(y^2) \), we find the partial derivative with respect to each variable separately.
This means:

• For \( \frac{\partial z}{\partial x} \): Treat \( y \) as a constant and differentiate with respect to \( x \).
• For \( \frac{\partial z}{\partial y} \): Treat \( x \) as a constant and differentiate with respect to \( y \).

The results are the building blocks for constructing the differential of a multivariable function. This process allows us to see the contribution of each variable to the rate of change of the function, which helps in understanding dynamics in fields like physics and engineering.

Multivariable Functions

Multivariable functions are those that depend on more than one variable. Here, \( z \) is a function dependent on \( x \) and \( y \). Such functions require special techniques for differentiation due to multiple interacting variables.
When differentiating:

• Consider how each variable affects the function separately through partial derivatives.
• Evaluate how changes in both variables influence the overall function \( z \).

Differentiating multivariable functions leads to expressions of differentials, such as \( dz = \left(\frac{\partial z}{\partial x} \right) dx + \left(\frac{\partial z}{\partial y}\right) dy \), that describe how smaller changes in \( x \) and \( y \) result in a change in \( z \).
Understanding these can help in a wide range of applications, from optimizing business processes to modeling complex systems.