Problem 36
Question
Find \(\partial f / 2 x, \partial f / \partial y\), and \(\partial f / \partial z\) for the given functions. . \(f(x, y, z)=\sin (x+y-z)\)
Step-by-Step Solution
Verified Answer
\(\partial f / \partial x = \cos(x + y - z)\), \(\partial f / \partial y = \cos(x + y - z)\), \(\partial f / \partial z = -\cos(x + y - z)\).
1Step 1: Understand the function
We have the function \(f(x, y, z) = \sin(x + y - z)\). Our task is to find the partial derivatives of this function with respect to \(x\), \(y\), and \(z\).
2Step 2: Partial derivative with respect to x
To find \(\frac{\partial f}{\partial x}\), treat \(y\) and \(z\) as constants and differentiate with respect to \(x\).The partial derivative is:\[ \frac{\partial f}{\partial x} = \cos(x + y - z) \cdot 1 = \cos(x + y - z) \]
3Step 3: Partial derivative with respect to y
To find \(\frac{\partial f}{\partial y}\), treat \(x\) and \(z\) as constants and differentiate with respect to \(y\).The partial derivative is:\[ \frac{\partial f}{\partial y} = \cos(x + y - z) \cdot 1 = \cos(x + y - z) \]
4Step 4: Partial derivative with respect to z
To find \(\frac{\partial f}{\partial z}\), treat \(x\) and \(y\) as constants and differentiate with respect to \(z\).The partial derivative is:\[ \frac{\partial f}{\partial z} = \cos(x + y - z) \cdot (-1) = -\cos(x + y - z) \]
Key Concepts
CalculusMultivariable FunctionsDifferentiation
Calculus
Calculus is an essential branch of mathematics that allows us to understand changes. It focuses on two main operations: differentiation and integration. Differentiation helps us find rates at which quantities change, while integration helps in summing infinitesimally small data points to find a whole.
In the context of functions, particularly those involving multiple variables, calculus assists in clarifying how changes in one variable affect the entire function. This becomes invaluable when analyzing dynamic systems, optimizing problems, and understanding complex relationships through functions that don't change linearly.
In this exercise, we are particularly interested in finding partial derivatives, which fall under differentiation. We explore how calculus enables us to systematically dissect each component in a multivariable function to better comprehend the overall behavior of the model.
In the context of functions, particularly those involving multiple variables, calculus assists in clarifying how changes in one variable affect the entire function. This becomes invaluable when analyzing dynamic systems, optimizing problems, and understanding complex relationships through functions that don't change linearly.
In this exercise, we are particularly interested in finding partial derivatives, which fall under differentiation. We explore how calculus enables us to systematically dissect each component in a multivariable function to better comprehend the overall behavior of the model.
Multivariable Functions
Multivariable functions are those which depend on more than one variable. For example, consider the function given in the exercise: \[f(x, y, z) = \sin(x + y - z)\] This function depends on three variables: \(x\), \(y\), and \(z\). Multivariable functions allow us to model complex systems that involve several varying conditions or dimensions simultaneously.
When dealing with multivariable functions, it's crucial to understand that each variable can independently influence the function's output. This is why understanding partial derivatives becomes so important; it gives us the tool we need to measure these changes individually. By holding the other variables constant, we can determine how changes in one variable specifically impact the result, allowing for in-depth analysis and more precise control over complex systems.
When dealing with multivariable functions, it's crucial to understand that each variable can independently influence the function's output. This is why understanding partial derivatives becomes so important; it gives us the tool we need to measure these changes individually. By holding the other variables constant, we can determine how changes in one variable specifically impact the result, allowing for in-depth analysis and more precise control over complex systems.
Differentiation
Differentiation is the process used to find out how a function changes as its input changes. With multivariable functions, differentiation becomes partial differentiation because we solve for each variable one at a time.
Let's revisit our function: \[f(x, y, z) = \sin(x + y - z)\] For this function, the differentiation process entails the following:
Let's revisit our function: \[f(x, y, z) = \sin(x + y - z)\] For this function, the differentiation process entails the following:
- ****Partial Derivative with respect to \(x\)**: We treat \(y\) and \(z\) as constants and differentiate the function with respect to \(x\). We use the derivative of \(\sin\) which is \(\cos\) to get: \(\frac{\partial f}{\partial x} = \cos(x + y - z)\).
- ****Partial Derivative with respect to \(y\)**: Similarly, holding \(x\) and \(z\) constant, differentiating with respect to \(y\) provides: \(\frac{\partial f}{\partial y} = \cos(x + y - z)\).
- ****Partial Derivative with respect to \(z\)**: This time, \(x\) and \(y\) are constants. The partial derivative becomes: \(\frac{\partial f}{\partial z} = -\cos(x + y - z)\), reversing the sign because of the negative influence of \(z\) in the argument.
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