Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(f(x, y, z)=e^{-x y z}-\ln \left(x y-z^{2}\right)\), find \(f_{x}(x, y, z)\)

Step-by-Step Solution

Answer

\(f_x(x, y, z) = -y z e^{-x y z} - \frac{y}{x y - z^2}\)

Step 1: Understand the Problem

We're asked to find the partial derivative of the function \(f(x, y, z) = e^{-x y z} - \ln(x y - z^2)\) with respect to the variable \(x\). This means we must differentiate \(f\) with respect to \(x\) while treating \(y\) and \(z\) as constants.

Step 2: Differentiate the Exponential Term

Consider the first term \(e^{-x y z}\). The derivative with respect to \(x\) involves the chain rule. Let \(u = -x y z\), then \(\frac{du}{dx} = -y z\) and \(\frac{d}{du}(e^u) = e^u\). Therefore, the derivative is \(\frac{d}{dx}(e^{-x y z}) = e^{-x y z} \cdot (-y z) = -y z e^{-x y z}\).

Step 3: Differentiate the Logarithmic Term

Now differentiate the second term \(-\ln(x y - z^2)\) with respect to \(x\). Recall the derivative of \(-\ln(u)\) is \(-\frac{1}{u} \cdot \frac{du}{dx}\). Here, \(u = x y - z^2\), so \(\frac{du}{dx} = y\). The derivative is therefore \(-\frac{y}{x y - z^2}\).

Step 4: Combine Results

Add the derivatives from Steps 2 and 3 to find the total derivative \(f_x(x, y, z)\). This gives:\[ f_x(x, y, z) = -y z e^{-x y z} - \frac{y}{x y - z^2} \]

Step 5: Simplify if Necessary

Observe the expression and confirm that it is already simplified. Thus, the partial derivative is:\[ f_x(x, y, z) = -y z e^{-x y z} - \frac{y}{x y - z^2} \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, particularly useful when dealing with composite functions. It allows us to differentiate a function that is nested with another function.
In our exercise, we encounter this with the expression \( e^{-xyz} \). Here, the function \( e^u \) has \( u = -xyz \). To apply the chain rule, we first differentiate the outer function with respect to the inner function, and then multiply by the derivative of the inner function with respect to the variable of interest.
In this situation, the steps look like this:

• Identify the inner function: \( u = -xyz \)
• Find the derivative of the outer function: \( \frac{d}{du}(e^u) = e^u \)
• Differentiate the inner function: \( \frac{du}{dx} = -yz \)

Putting these together gives the result for the derivative: \( \frac{d}{dx}(e^{-xyz}) = e^{-xyz} \cdot (-yz) = -yze^{-xyz} \). By following these steps, we can clearly see how the chain rule simplifies the differentiation process for exponential functions with a composite argument.

Exponential Functions

Exponential functions are characterized by a constant base raised to the power of a variable. In calculus, they often show up in processes that grow or decay at a rate proportional to their current value.
The general derivative rule of an exponential function \( e^u \) is simple and elegant because the base, \( e \), is special due to its natural properties.

• The derivative of \( e^u \) will always be \( e^u \times \frac{du}{dx} \), which makes it closely related to its own output.
• For any function \( u(x) \), \( \frac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot \frac{du}{dx} \). This can be directly noticed in the exercise for \( e^{-xyz} \).

The ease of differentiating exponential functions is one reason why they are prevalent in mathematical models across various scientific fields. They provide straightforward solutions, since their derivatives maintain the form of the original function, modified by the chain rule if needed.

Logarithmic Derivatives

Differentiating logarithmic functions involves knowing the rules specific to logs, such as the natural logarithm, \( \ln(x) \). Logarithmic derivatives prove very useful because they simplify multiplication and division problems into addition and subtraction.
For a natural logarithmic function, its derivative, \( \ln(u) \), is computed by using the formula:

• Base rule: \( \frac{d}{dx} [\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} \).
• Don't forget the minus in front when required, for instance, \( -\ln(x y - z^2)\) becomes \( -\frac{y}{x y - z^2} \) as demonstrated in our solution.

In practical terms, by applying the rule for the derivative of a logarithmic term, we simplify the complexity of expressions involving such differentiations. The ability to handle intricate expressions through straightforward differentiation rules enhances our capability to solve a variety of mathematical problems, particularly those involving exponential growth or decay models.