Problem 48 - Calculus Early Transcendentals

Question

Use implicit differentiation to find \(\partial z / \partial x\) and \(\partial z / \partial y\) $$\sin (x y z)=x+2 y+3 z$$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial x} = \frac{1 - \cos(xyz) yz}{\cos(xyz) xy - 3} \) and \( \frac{\partial z}{\partial y} = \frac{2 - \cos(xyz) xz}{\cos(xyz) xy - 3} \).

Step 1: Differentiate with respect to x

Start by differentiating both sides of the equation with respect to \( x \). Since \( z \) is a function of \( x \) and \( y \), use implicit differentiation. The left side becomes \( \cos(xyz)(yz + xy \frac{\partial z}{\partial x}) \) and the right side differentiates to \( 1 + 3 \frac{\partial z}{\partial x} \).

Step 2: Solve for \( \frac{\partial z}{\partial x} \)

Equate the differentiated expressions from Step 1: \[ \cos(xyz)(yz + xy \frac{\partial z}{\partial x}) = 1 + 3 \frac{\partial z}{\partial x} \]Rearrange the terms to solve for \( \frac{\partial z}{\partial x} \):\[ \cos(xyz) yz + \cos(xyz) xy \frac{\partial z}{\partial x} = 1 + 3 \frac{\partial z}{\partial x} \]\[ \cos(xyz) xy \frac{\partial z}{\partial x} - 3 \frac{\partial z}{\partial x} = 1 - \cos(xyz) yz \]Factor out \( \frac{\partial z}{\partial x} \):\[ \frac{\partial z}{\partial x}(\cos(xyz) xy - 3) = 1 - \cos(xyz) yz \]\[ \frac{\partial z}{\partial x} = \frac{1 - \cos(xyz) yz}{\cos(xyz) xy - 3} \]

Step 3: Differentiate with respect to y

Differentiate both sides of the equation with respect to \( y \). For the left side, using implicit differentiation, it becomes \( \cos(xyz)(xz + xy \frac{\partial z}{\partial y}) \) and the right side gives \( 2 + 3 \frac{\partial z}{\partial y} \).

Step 4: Solve for \( \frac{\partial z}{\partial y} \)

Set the differentiated expressions equal from Step 3:\[ \cos(xyz)(xz + xy \frac{\partial z}{\partial y}) = 2 + 3 \frac{\partial z}{\partial y} \]Rearrange to isolate \( \frac{\partial z}{\partial y} \):\[ \cos(xyz) xz + \cos(xyz) xy \frac{\partial z}{\partial y} = 2 + 3 \frac{\partial z}{\partial y} \]\[ \cos(xyz) xy \frac{\partial z}{\partial y} - 3 \frac{\partial z}{\partial y} = 2 - \cos(xyz) xz \]Factor out \( \frac{\partial z}{\partial y} \):\[ \frac{\partial z}{\partial y}(\cos(xyz) xy - 3) = 2 - \cos(xyz) xz \]\[ \frac{\partial z}{\partial y} = \frac{2 - \cos(xyz) xz}{\cos(xyz) xy - 3} \]

Key Concepts

Partial Derivatives

Partial derivatives are a core concept in multivariable calculus. They allow us to focus on how a function changes with respect to one variable, while keeping the others constant. This is crucial for understanding functions with more than one variable, like the one in our exercise. The function given is \( \sin(xyz) = x + 2y + 3z \). Here, we treat \( z \) as a function of \( x \) and \( y \).
To find \( \partial z / \partial x \), we differentiate the equation with respect to \( x \), resulting in both \( z \) and \( y \) being considered as constants for partial differentiation, except where \( z \) is explicitly dependent. This gives us a differential expression that relates changes in \( z \) to changes in \( x \).

• For \( \partial z/\partial x \) the equation was differentiated with respect to \( x \) and simplified accordingly.
• Then, to isolate \( \partial z/\partial x \), algebraic manipulation and factorization were used.

Similarly, to find \( \partial z / \partial y \), the function is differentiated with respect to \( y \), treating \( x \) as constant. The process results in another equation that shows how \( z \) changes with \( y \). By solving these equations, we get the partial derivatives.

Trigonometric Functions

Trigonometric functions like sine and cosine are pervasive in calculus, especially when dealing with curves and oscillations. In the given exercise, the equation \( \sin(xyz) = x + 2y + 3z \) prominently features the sine function. When differentiating such functions, it becomes essential to correctly apply the chain rule.
The chain rule allows us to differentiate composite functions by taking the derivative of the outer function and multiplying it by the derivative of the inner function. In this case, differentiating \( \sin(xyz) \) gives us \( \cos(xyz) \times (d/dx(xyz)) \).

• Understanding sin and cos functions helps predict how the graph of the equation behaves.
• They are cyclic and periodic, which means they repeat patterns over given intervals.

When dealing with trigonometric functions in multivariable contexts, remember that the differentiation is akin to pulling apart layers and requires patience to perform accurately.

Multivariable Calculus

In multivariable calculus, we deal with functions of more than one variable, providing a broader scope than single-variable calculus. This branch of calculus extends the concepts of differentiation and integration into higher dimensions. For our problem, we must understand how \( x \), \( y \), and \( z \) interact within the equation \( \sin(xyz) = x + 2y + 3z \).
Multivariable calculus also involves learning new concepts important for analyzing multi-dimensional systems:

• **Multivariable functions:** Functions that have two or more variables, like \( f(x, y, z) \).
• **Implicit differentiation:** Often required when functions are not easily separated into simpler parts.
• **Partial derivatives:** Used to measure the rate of change of a function concerning one variable while holding others fixed.

By applying these concepts, we can determine how each variable influences another, shedding light on the dynamic interactions of complex systems. This helps model real-world scenarios where multiple factors impact outcomes simultaneously.