Problem 32 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$w=z e^{x y z}$$

Step-by-Step Solution

Answer

The partial derivatives are: \(\frac{\partial w}{\partial x} = yz^2 e^{xyz}\), \(\frac{\partial w}{\partial y} = xz^2 e^{xyz}\), \(\frac{\partial w}{\partial z} = (1 + xy) ze^{xyz}\).

Step 1: Understanding the Partial Derivatives

To find the partial derivatives of the function, we need to calculate how the function changes with each of its variables while keeping the other variables constant. For this function, we will find the partial derivatives with respect to \(x\), \(y\), and \(z\).

Step 1: Calculate \(\frac{\partial w}{\partial x}\)

Treat \(y\) and \(z\) as constants. Notice that the term \(e^{xyz}\) is dependent on \(x\), so its derivative with respect to \(x\) needs the chain rule. The differentiation of \(e^{xyz}\) is \(yz \cdot e^{xyz}\). Therefore, the partial derivative is:\[ \frac{\partial w}{\partial x} = z \cdot yz \cdot e^{xyz} = yz^2 e^{xyz}.\]

Step 2: Calculate \(\frac{\partial w}{\partial y}\)

Similarly, treat \(x\) and \(z\) as constants. The derivative of \(e^{xyz}\) with respect to \(y\) is \(xz \cdot e^{xyz}\). Therefore, the partial derivative is:\[ \frac{\partial w}{\partial y} = z \cdot xz \cdot e^{xyz} = xz^2 e^{xyz}.\]

Step 3: Calculate \(\frac{\partial w}{\partial z}\)

Now treat \(x\) and \(y\) as constants. This problem has two terms for \(w\): \(z\) itself and \(e^{xyz}\). - The derivative of \(z\) with respect to \(z\) is \(1\).- The derivative of \(e^{xyz}\) is \(xy \cdot e^{xyz}\), using the chain rule.Combining these results, we obtain:\[\frac{\partial w}{\partial z} = \frac{\partial}{\partial z} (z \cdot e^{xyz}) = 1 \cdot e^{xyz} + z \cdot xy \cdot e^{xyz} = e^{xyz} + xyz e^{xyz} = (1 + xy) ze^{xyz}.\]

Key Concepts

Chain Rule in Partial Derivatives

When diving into partial derivatives, the chain rule is an indispensable tool. It helps us differentiate complex functions that are compositions of other functions. In the context of multivariable calculus, the chain rule comes into play when a function depends on more than one variable, and we have terms like \(e^{xyz}\). Here, each variable is potentially interacting with the others.
In the function \(w = z e^{xyz}\), to find partial derivatives, we differentiate with respect to one variable at a time, treating the others as constants. The chain rule allows us to correctly determine the derivative of nested functions.
Using the chain rule:

• For \(\frac{\partial w}{\partial x}\), the chain rule gives \( yz \cdot e^{xyz} \) because \(xyz\) acts as a single unit affected by changes in \(x\).
• Similarly, \(\frac{\partial w}{\partial y}\) makes use of the chain rule to get \( xz \cdot e^{xyz} \).
• Finally, \(\frac{\partial w}{\partial z}\) requires using the derivative of \(z\) itself and \(e^{xyz}\), adding a layer of complexity that the chain rule helps unravel.

Grasping the chain rule's role is crucial in finding correct partial derivatives when dealing with exponential and trigonometric functions within multivariable settings.

Multivariable Calculus and Partial Derivatives

Multivariable calculus is an extension of calculus into more than one variable. It allows us to explore how functions change in space, taking us beyond single-variable scenarios into a realm where several factors are at play. Partial derivatives are key components of this field. They let us analyze how a function behaves along each variable’s axis independently.
In our function \(w = z e^{xyz}\), finding partial derivatives involves isolating the change in \(w\) as one variable shifts while others remain fixed. This perspective is pivotal:

• For \(\frac{\partial w}{\partial x}\): Treats \(y\) and \(z\) as constants, capturing how \(w\) adjusts with shifts in \(x\).
• For \(\frac{\partial w}{\partial y}\): Fixes \(x\) and \(z\) to see changes along the \(y\)-axis only.
• For \(\frac{\partial w}{\partial z}\): Observes the alteration in \(w\) when \(z\) is altered, keeping \(x\) and \(y\) constant.

Understanding these independent shifts provides insights into how a multidimensional function behaves. Multivariable calculus opens doors to fields like engineering and physics, where dynamics depend on multiple, changing factors.

Exponential Functions in Multivariable Calculus

Exponential functions, like \(e^{xyz}\), are powerhouses in mathematics due to their smooth and consistent rate of change. In multivariable calculus, they introduce interesting complexities, especially within contexts requiring partial derivatives.
Exponential expressions adapt swiftly to changes in their base, which here is the product \(xyz\). Each variable's influence on the base leads to specific derivative outcomes:

• The exponential function \(e^{xyz}\) becomes critical since any slight alteration in \(x\), \(y\), or \(z\) dramatically impacts \(w\), outlined by derivatives \(yz e^{xyz}\) or \(xz e^{xyz}\).
• The unchanging nature of exponential growth is integral to understanding the predictable yet complex behavior of \(w\).
• We leverage the exponential rule in the partial derivative calculation, ensuring smooth transitions between variables in multivariable calculus.

These properties make exponential functions invaluable in scientific applications where exponential decay or growth models describe real-world phenomena, such as population dynamics or radioactive decay.