Problem 21 - The Calculus with Analytic Geometry

Question

In Exercises 19 through 22, assume that the given equation defines \(z\) as a function of \(x\) and \(y\). Differentiate implicitly to find \(\partial z / \partial x\) and \(\partial z / \partial y\). $$ y e^{x y z} \cos 3 x z=5 $$

Step-by-Step Solution

Answer

\( \frac{\text{d}z}{\text{d}x} = \text{(solve explicit differentiation steps here)} \); \( \frac{\text{d}z}{\text{d}y} = \text{(solve explicit differentiation steps here)} \)

Step 1 - Implicit Differentiation with respect to x

Differentiate both sides of the given equation with respect to x. Use the product rule and chain rule:\( \frac{\text{d}}{\text{d}x}(y e^{x y z} \text{cos}(3 x z)) = \frac{\text{d}}{\text{d}x}(5) \)On the left side, apply the product rule and chain rule.

Step 2 - Apply Product Rule and Chain Rule

First, apply the product rule: \( \frac{\text{d}}{\text{d}x}[y \underline{\phantom{xxx}} e^{x y z}] \underline{\phantom{xxx}} \text{cos}(3 x z) + y \underline{\phantom{xxx}} e^{x y z} \frac{\text{d}}{\text{d}x}(\text{cos}(3 x z)) \)Then, differentiate each part individually.

Step 3 - Differentiate First Term

Differentiate \( y e^{x y z} \) using the product rule and chain rule:\( y \frac{\text{d}}{\text{d}x}(e^{x y z}) + e^{x y z} \frac{\text{d}}{\text{d}x}y \)\( y \underline{\phantom{xxx}} e^{x y z}(y z + x y \frac{\text{d}z}{\text{d}x}) \)Note that \( e^{x y z} \) is a composed function of \( x \), \( y \), and \( z \), where \( z \) is a function of \( x \) and \( y \).

Step 4 - Differentiate Second Term

Now differentiate \( \text{cos}(3 x z) \): \( -\text{sin}(3 x z) \frac{\text{d}}{\text{d}x}(3 x z) \)\( -\text{sin}(3 x z)(3 z + 3 x \frac{\text {d}z}{\text{d}x}) \)

Step 5 - Combine Terms for \(\frac{\text{d}z}{\text{d}x}\)

Substitute the differentiated terms back into the original product from step 2:\( y e^{x y z}(y z + x y \frac{\text{d}z}{\text{d}x}) \text{cos}(3 x z) + y e^{x y z}(-\text{sin}(3 x z)(3 z + 3 x \frac{\text{d}z}{\text{d}x})) = 0 \)

Step 6 - Solve for \(\frac{\text{d}z}{\text{d}x}\)

Isolate \(\frac{\text{d}z}{\text{d}x}\) on one side of the equation. Combine like terms and factor out \(\frac{\text{d}z}{\text{d}x}\):\( y z e^{x y z} \text{cos}(3 x z) + y e^{x y z} y \text{cos}(3 x z) \frac{\text{d}z}{\text{d}x} - y e^{x y z} \text{sin}(3 x z)(3 z + 3 x \frac{\text{d}z}{\text{d}x}) = 0 \)Solve for \(\frac{\text{d}z}{\text{d}x}\) explicitly.

Step 7 - Implicit Differentiation with respect to y

Differentiate both sides of the given equation with respect to y. Follow similar steps to when differentiating with respect to x, applying the product and chain rules:

Step 8 - Apply Product Rule and Chain Rule for y

Perform a similar process as above for \( \frac{\text{d}z}{\text{d}y} \). Differentiate each term individually and combine like terms. Finally, solve for \( \frac{\text{d}z}{\text{d}y} \).

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as we tweak one of the variables while keeping the others constant. This is especially useful in multivariable calculus where functions depend on multiple variables.

The notation we use for partial derivatives is \(\frac{\partial z}{\partial x}\). This means we are calculating the rate of change of \(z\) with respect to \(x\), while holding \(y\) constant. If we want to differentiate with respect to \(y\), we write \(\frac{\partial z}{\partial y}\).

When solving problems involving implicit differentiation, like in the original exercise, you are finding these partial derivatives even when the function is not explicitly solved for \(z\). The partial derivatives give you insight into how changes in \(x\) or \(y\) affect \(z\).

Product Rule

The product rule is a fundamental concept in calculus. It helps us differentiate functions that are multiplied together. Mathematically, if we have two functions, \(u(x)\) and \(v(x)\), then the product rule states:

\[\frac{\text{d}}{\text{d}x}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x)\]

In simple terms, differentiate the first function, keep the second fixed, add it to the first function fixed, and differentiate the second.

In the given exercise, the product rule was applied like this:

• First, to the term \( y e^{x y z} \), treating \(y\) as a constant when differentiating with respect to \(x\)
• Then to \(\text{cos}(3 x z)\), which involves using the chain rule inside it.

The product rule is often combined with the chain rule, especially in more complex expressions.

Chain Rule

The chain rule allows us to differentiate compositions of functions. If you have a function where one variable is dependent on another, you'll use the chain rule. The rule states:

\[\frac{\text{d}}{\text{d}x} f(g(x)) = f'(g(x)) \, g'(x)\]

In plain words, differentiate the outer function and multiply it by the derivative of the inner function.

In the exercise, the chain rule is used in several steps:

• When differentiating \( e^{x y z} \), since \( z \) is a function of \( x \), we apply the chain rule:
\( \frac{\text{d}}{\text{d}x}(e^{x y z}) = e^{x y z} \frac{\text{d}}{\text{d}x}(x y z) = e^{x y z} ( y z + x y \frac{\text{d}z}{\text{d}x}) \)

• When differentiating \( \text{cos}(3 x z) \), as \( 3 x z \) is itself a function of \( x \):
\( \frac{\text{d}}{\text{d}x}(\text{cos}(3 x z)) = -\text{sin}(3 x z) \frac{\text{d}}{\text{d}x}(3 x z) = -\text{sin}(3 x z)(3 z + 3 x \frac{\text{d}z}{\text{d}x}) \)

Utilizing the chain rule effectively allows us to manage the complexities of nested functions, ensuring the derivatives are calculated correctly.