Problem 22 - 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 \(\partial z / \partial y\). $$ z e^{y z}+2 x e^{x z}-4 e^{x y}=3 $$

Step-by-Step Solution

Answer

The partial derivatives are \( \frac{\partial z}{\partial x} = \frac{4y e^{xy} - 2 e^{xz}}{e^{yz} + yz e^{yz} + 2x e^{xz} z} \) and \( \frac{\partial z}{\partial y} = \frac{4x e^{xy} - e^{yz}}{z e^{yz} + yz z e^{yz} + 2x z e^{xz}} \).

Step 1 - Differentiate Implicitly with Respect to x

Differentiate each term of the equation with respect to x. Apply the product rule and chain rule where necessary: \( \frac{\partial}{\partial x} \left( z e^{yz} \right) + \frac{\partial}{\partial x} \, \left( 2x e^{xz} \right) - \frac{\partial}{\partial x} \, \left( 4 e^{xy} \right) = \frac{\partial}{\partial x} 3 \). For each term: \( \frac{\partial}{\partial x} \left( z e^{yz} \right) = e^{yz} \frac{\partial z}{\partial x} + yz e^{yz} \frac{\partial z}{\partial x} \) using the product rule.\( \frac{\partial}{\partial x} \, \left( 2x e^{xz} \right) = 2 e^{xz} + 2x e^{xz} z \frac{\partial z}{\partial x} \) using the product rule. \( \frac{\partial}{\partial x} \, \left( 4 e^{xy} \right) = 4 y e^{xy} \).

Step 2 - Simplify and Solve for \( \partial z / \partial x \)

Combine the differentiated terms and simplify: \( e^{yz} \frac{\partial z}{\partial x} + yz e^{yz} \frac{\partial z}{\partial x} + 2 e^{xz} + 2x e^{xz} z \frac{\partial z}{\partial x} - 4 y e^{xy} = 0 \). Factor out \( \frac{\partial z}{\partial x} \): \( \left( e^{yz} + yz e^{yz} + 2x e^{xz} z \right) \frac{\partial z}{\partial x} = -2 e^{xz} + 4 y e^{xy} \). Solve for \( \frac{\partial z}{\partial x} \): \( \frac{\partial z}{\partial x} = \frac{4 y e^{xy} - 2 e^{xz}}{e^{yz} + yz e^{yz} + 2x e^{xz} z} \).

Step 3 - Differentiate Implicitly with Respect to y

Differentiate each term of the original equation with respect to y. Apply the product rule and chain rule where necessary: \( \frac{\partial}{\partial y} \left( z e^{yz} \right) + \frac{\partial}{\partial y} \, \left( 2x e^{xz} \right) - \frac{\partial}{\partial y} \, \left( 4 e^{xy} \right) = \frac{\partial}{\partial y} 3 \). For each term: \( \frac{\partial}{\partial y} \left( z e^{yz} \right) = e^{yz} z \frac{\partial z}{\partial y} + yz e^{yz} z \frac{\partial z}{\partial y} + e^{yz} \). \( \frac{\partial}{\partial y} \, \left( 2x e^{xz} \right) = 2x z e^{xz} \frac{\partial z}{\partial y} \). \( \frac{\partial}{\partial y} \, \left( 4 e^{xy} \right) = 4x e^{xy} \).

Step 4 - Simplify and Solve for \( \partial z / \partial y \)

Combine and simplify: \( z e^{yz} \frac{\partial z}{\partial y} + yz z e^{yz} \frac{\partial z}{\partial y} + e^{yz} + 2x z e^{xz} \frac{\partial z}{\partial y} - 4x e^{xy} = 0 \). Factor out \( \frac{\partial z}{\partial y} \): \( \left( z e^{yz} + yz z e^{yz} + 2x z e^{xz} \right) \frac{\partial z}{\partial y} = 4x e^{xy} - e^{yz} \). Solve for \( \frac{\partial z}{\partial y} \): \( \frac{\partial z}{\partial y} = \frac{4x e^{xy} - e^{yz}}{z e^{yz} + yz z e^{yz} + 2x z e^{xz}} \).

Key Concepts

partial derivatives

Partial derivatives refer to the derivatives of a function with respect to one variable while keeping the other variables constant. When working with multivariable functions, partial derivatives are essential. Let's look at an example to understand this better. Consider the function given in the exercise:

$$ z e^{yz} + 2 x e^{xz} - 4 e^{xy} = 3 $$
The partial derivative with respect to x (\frac{\frac{\partial}}{\partial x}}) means we treat y as a constant. Similarly, the partial derivative with respect to y (\frac{\frac{\partial}}{\partial y}}) treats x as a constant. When we differentiate implicitly, it's essential to apply the differentiation rules carefully, especially when dealing with products and compositions of functions, which brings us to two crucial differentiation rules: the product rule and chain rule.

product rule

The product rule is essential when differentiating a product of two or more functions. It states:

$$( u \times v )' = u' \times v + u \times v' $$
This means we differentiate the first function and multiply it by the second function, then add the first function multiplied by the derivative of the second function.

In the provided exercise, we use the product rule multiple times. For instance, for the term $$ 2 x e^{xz} $$

The differentiation would be:

\frac{\frac{\partial}{\partial x}} ( 2 x e^{xz} ) = 2' e^{xz} + 2 x \frac{\frac{\partial}{\frac{{\round{x z}}}}}{∂x z} => 2 e^{xz}+ 2 x e^{xz}z \frac{\frac{\partial z}{ \frac{{❑ x}}}} Using this rule correctly helps us in handling differentiated products efficiently.

chain rule

The chain rule is helpful when differentiating composite functions. It states that if a function y is defined as a function of u, which is a function of x, then:

$${ dy \frac{{\frac{\partial y}{\frac{\round ∂çə \frac{\round}}{{\frac{\round}}}} \frac{{$$}} ))$$ \frac{\frac{dz}{δ}}