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Use the result of Exercise 48 to evaluate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for the following relations. $$x y z+x+y-z=0$$

Short Answer

Expert verified
Question: Find the partial derivatives of \(z\) with respect to \(x\) and \(y\) for the given relation \(xyz + x + y - z = 0\). Answer: The partial derivatives of \(z\) are \(\frac{\partial z}{\partial x} = yz + 1\) and \(\frac{\partial z}{\partial y} = xz + 1\).

Step by step solution

01

Differentiate both sides with respect to x

First, we differentiate both sides of the equation with respect to \(x\). Remember to treat \(y\) and \(z\) as constants during differentiation: $$\frac{\partial}{\partial x} (xyz + x + y - z) = \frac{\partial}{\partial x} (0)$$
02

Compute the result of differentiation

Compute the left side of the equation: $$\frac{\partial}{\partial x}(xyz) + \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial x}(y) - \frac{\partial}{\partial x}(z)$$ We get: $$yz + 1 - 0 - \frac{\partial z}{\partial x} = 0$$
03

Solve for the first partial derivative

Now, we can solve for \(\frac{\partial z}{\partial x}\) from the equation above: $$\frac{\partial z}{\partial x} = yz + 1$$
04

Differentiate both sides with respect to y

Next, we need to differentiate both sides of the equation with respect to \(y\). Remember to treat \(x\) and \(z\) as constants during differentiation: $$\frac{\partial}{\partial y} (xyz + x + y - z) = \frac{\partial}{\partial y} (0)$$
05

Compute the result of differentiation

Compute the left side of the equation: $$\frac{\partial}{\partial y}(xyz) + \frac{\partial}{\partial y}(x) + \frac{\partial}{\partial y}(y) - \frac{\partial}{\partial y}(z)$$ We get: $$xz + 0 + 1 - \frac{\partial z}{\partial y} = 0$$
06

Solve for the second partial derivative

Now, we can solve for \(\frac{\partial z}{\partial y}\) from the equation above: $$\frac{\partial z}{\partial y} = xz + 1$$
07

Write the final answer

We have found the partial derivatives of \(z\) with respect to \(x\) and \(y\): $$\frac{\partial z}{\partial x} = yz + 1$$ $$\frac{\partial z}{\partial y} = xz + 1$$

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Key Concepts

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Multivariable Calculus
Multivariable calculus is an extension of calculus that involves functions with more than one variable. It can seem daunting because it builds upon the principles you already know from single-variable calculus but extends them into higher dimensions. In multivariable calculus, we often study functions like \( f(x, y, z) \), which depend on several variables and are represented in a multi-dimensional space. A primary focus in this area is understanding how changes in one variable affect the others, which is where concepts like partial derivatives come into play.Partial derivatives help us explore rates of change in multivariable functions. We can think of them as seeing the change in a function when only one of its variables changes, much like tracing a path along a single axis on a graph while keeping all other paths fixed. This is crucial for describing dynamic, real-world scenarios where many factors are interconnected.
  • Partial differentiation lets us isolate the effect of one variable at a time.
  • This process provides insight into how different inputs interact in complex systems.
By harnessing the power of multivariable calculus, we can more deeply understand systems in physics, engineering, economics, and beyond.
Chain Rule
The chain rule is a vital concept in calculus that allows us to differentiate composite functions. In the context of multivariable calculus, the chain rule becomes more complex, but it remains a powerful tool for solving differentiation problems that involve multiple interdependent variables.In its simplest form, the chain rule states that if you have a function composed of other functions, like \( f(g(x)) \), the derivative of \( f \) with respect to \( x \) is the derivative of \( f \) with respect to \( g \) times the derivative of \( g \) with respect to \( x \). For multivariable functions, the chain rule expands to account for changes across more dimensions, leading to expressions that involve the partial derivatives of the intermediate variables.
  • It helps in taking derivatives of nested functions or functions within functions.
  • The chain rule ensures that all pathways of variable interaction are accounted for when differentiating.
This approach is particularly useful when functions like \( z(x, y) \) depend on multiple other variables, requiring derivatives such as \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).When employing the chain rule, it is essential to carefully track each link of dependency, especially in complex equations where each variable might independently and jointly influence each other.
Implicit Differentiation
Implicit differentiation is a technique used when dealing with equations that define one variable in terms of another without explicitly being solved for that variable. This method is particularly beneficial when the relationship between variables is complicated enough that isolating one variable is difficult or impossible.In implicit differentiation, we differentiate each term of the equation with respect to a particular variable, treating other variables as constants temporarily. This is akin to tracing the web of dependencies that weave through multivariable functions, allowing us to understand how one variable implicitly influences others, much like finding that \( z \) is a function of \( x \) and \( y \) without solving directly for \( z \).
  • Used when it's hard to express one variable explicitly in terms of others.
  • Helps derive derivatives when dealing with complex relationships.
  • Maintains the integrity of interconnected variable relationships, honoring constraints imposed by equations.
In our exercise, when examining the equation \( xyx + x + y - z = 0 \), implicit differentiation allows us to determine \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) by differentiating the whole equation step by step, thus preserving the inherent complexity of the original relationship.

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