Problem 30 - Advanced Engineering Mathematics

Question

Find the first partial derivatives of the given function. $$ w=x y \ln (x z) $$

Step-by-Step Solution

Answer

Partial derivatives: \( \frac{\partial w}{\partial x} = y \ln(xz) + y \), \( \frac{\partial w}{\partial y} = x \ln(xz) \), \( \frac{\partial w}{\partial z} = \frac{xy}{z} \).

Step 1: Find the Partial Derivative with respect to x

To find \( \frac{\partial w}{\partial x} \), treat \( y \) and \( z \) as constants. The function can be rewritten as \( w = xy \ln(xz) \). Using the product rule and derivative of \( \ln(u) \), where \( u = xz \), we have:\[ \frac{dw}{dx} = y \left( \ln(xz) + \frac{xz}{xz} \right) = y \ln(xz) + y\]

Step 2: Find the Partial Derivative with respect to y

To find \( \frac{\partial w}{\partial y} \), treat \( x \) and \( z \) as constants. The function becomes \( xy \ln(xz) \). The partial derivative is straightforward as \( x \ln(xz) \) is constant with respect to \( y \):\[ \frac{\partial w}{\partial y} = x \ln(xz)\]

Step 3: Find the Partial Derivative with respect to z

For the partial derivative \( \frac{\partial w}{\partial z} \), treat \( x \) and \( y \) as constants. Rewrite the function as \( w = xy \ln(xz) \). Using the chain rule, where derivative of \( \ln(u) = \frac{1}{u} \) and \( u = xz \):\[ \frac{\partial w}{\partial z} = xy \cdot \frac{1}{xz} \cdot x = \frac{xy}{z}\]

Key Concepts

Product Rule

When dealing with functions that are products, especially in calculus, the product rule is a powerful tool that ensures we differentiate accurately. It's important when you're finding derivatives of functions where two factors—let's call them \( f(x) \) and \( g(x) \)—are multiplied. The product rule states that

• \( (fg)' = f'g + fg' \)

In our exercise, consider the function as \( w = x \, y \, \ln(xz) \), making it imperative to utilize this rule.
Treat \( y \ln(xz) \) as one function and \( x \) as the other. When differentiating with respect to \( x \), the product rule helps us break down the problem into manageable pieces. It's crucial here to also apply other derivative rules, like differentiating the natural log function.

Chain Rule

The chain rule is essential when you have composite functions, that is, a function nested within another function. Essentially, it allows us to differentiate a composite function and is expressed as:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

In the exercise, this comes into play when dealing with the term \( \ln(xz) \). Here, \( u = xz \) is a function nested inside another (the natural logarithm), and differentiating \( \ln(u) \) involves first treating \( u \) as a variable, then finding the derivative of \( u \) with respect to \( z \).
This step, where we multiply the derivative of the outer function by the derivative of the inner function, ensures accurate differentiation.

Logarithmic Differentiation

Logarithmic differentiation can simplify differentiating complex products or quotients. It uses properties of logarithms, such as \( \ln(ab) = \ln a + \ln b \), which can be beneficial when multiple variables or terms are involved in a function.
In this exercise, the function \( w = xy \ln(xz) \) uses logarithmic differentiation particularly when working with \( \ln(xz) \). This interaction reduces the complexity of the expression because taking the logarithm of a product simplifies it into a sum, which is often easier to differentiate.

• For example, \( \ln(xz) = \ln x + \ln z \)

This approach allows us to apply basic differentiation rules to simplified components of the function.

Multivariable Calculus

Multivariable calculus extends basic calculus concepts to functions with more than one variable. This branch of mathematics involves techniques like partial differentiation, where each variable in a function is treated separately as others remain constant.
The exercise illustrates multivariable calculus as we find partial derivatives—\( \frac{\partial w}{\partial x} \), \( \frac{\partial w}{\partial y} \), and \( \frac{\partial w}{\partial z} \)—of the function \( w = xy \ln(xz) \).

• Each partial derivative provides the rate of change of the function with respect to one variable at a time.
• This helps in constructing a broader view of the function’s behaviour in higher dimensions.

Mastering partial derivatives equips you with vital tools to solve real-world problems involving multiple changing conditions.