Chapter 12: Problem 9
Find all first partial derivatives of each function. \(g(x, y)=e^{-x y}\)
Short Answer
Expert verified
The partial derivatives are \(g_x = -y e^{-xy}\) and \(g_y = -x e^{-xy}\).
Step by step solution
01
Identify Function and Variables
The function given is \(g(x, y) = e^{-xy}\). Here, \(x\) and \(y\) are the variables with respect to which we need to find the partial derivatives.
02
Find the Partial Derivative with Respect to x
To find the partial derivative of \(g\) with respect to \(x\), consider \(y\) as a constant. Apply the chain rule: the derivative of \(e^{-xy}\) with respect to \(x\) is \(-y e^{-xy}\). This gives us: \(g_x = -y e^{-xy}\).
03
Find the Partial Derivative with Respect to y
To find the partial derivative of \(g\) with respect to \(y\), consider \(x\) as a constant. Using the chain rule again, the derivative of \(e^{-xy}\) with respect to \(y\) is \(-x e^{-xy}\). This derives: \(g_y = -x e^{-xy}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental tool in calculus, particularly useful when dealing with functions composed of other functions. In simple terms, it helps us find the derivative of a composite function. Here's a quick way to think about it:
If you have an outer function and an inner function, the chain rule tells you to first find the derivative of the outer function, leaving the inner function unchanged. Then, multiply it by the derivative of the inner function.
In the given example, function: \( g(x, y) = e^{-xy} \), the outer function is the exponential \( e^u \) and the inner function is \( u = -xy \). To find the derivative, differentiate \( e^u \) (which remains \( e^u \), as the derivative of an exponential is itself), and multiply it by the derivative of \( -xy \) with respect to the chosen variable (either \( x \) or \( y \)).
This results in partial derivatives with respect to \( x \) and \( y \). Understanding the chain rule is crucial in multivariable calculus.
If you have an outer function and an inner function, the chain rule tells you to first find the derivative of the outer function, leaving the inner function unchanged. Then, multiply it by the derivative of the inner function.
In the given example, function: \( g(x, y) = e^{-xy} \), the outer function is the exponential \( e^u \) and the inner function is \( u = -xy \). To find the derivative, differentiate \( e^u \) (which remains \( e^u \), as the derivative of an exponential is itself), and multiply it by the derivative of \( -xy \) with respect to the chosen variable (either \( x \) or \( y \)).
This results in partial derivatives with respect to \( x \) and \( y \). Understanding the chain rule is crucial in multivariable calculus.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions with more than one variable. Here, instead of focusing solely on one input, you work with functions that depend on multiple inputs.
Think of multivariable calculus as exploring surfaces or multidimensional spaces, as opposed to simple curves. This means considering partial derivatives, gradients, and more complex integrations.
In our example, we have two variables, \( x \) and \( y \), each affecting \( g(x, y) \). This interplay between \( x \) and \( y \) adds depth to the solution, as we need to understand how each variable influences the overall function.
The importance of multivariable calculus lies in its applications - from physics to economics. Many real-world phenomena are best captured when we consider how several variables interact. By mastering partial derivatives and chain rule in this context, we can dive deeper into these analyses.
Think of multivariable calculus as exploring surfaces or multidimensional spaces, as opposed to simple curves. This means considering partial derivatives, gradients, and more complex integrations.
In our example, we have two variables, \( x \) and \( y \), each affecting \( g(x, y) \). This interplay between \( x \) and \( y \) adds depth to the solution, as we need to understand how each variable influences the overall function.
The importance of multivariable calculus lies in its applications - from physics to economics. Many real-world phenomena are best captured when we consider how several variables interact. By mastering partial derivatives and chain rule in this context, we can dive deeper into these analyses.
Partial Derivatives
Partial derivatives focus on how a function changes as one specific variable changes while keeping the others constant. When dealing with multivariable functions, they become essential tools.
To find a partial derivative, select a "variable of interest" and temporarily treat all other variables as constants. This allows you to examine the influence of just one variable at a time.
In the original exercise, finding \( g_x \) and \( g_y \) (the partial derivatives of \( g \) with respect to \( x \) and \( y \) respectively) required us to treat \( y \) as a constant for \( g_x \) and \( x \) as a constant for \( g_y \).
The results provided insight into how changes in one variable affect the function, holding the others steady. This is particularly useful when you need to determine sensitivity or rate of change in one specific direction within a multidimensional space.
To find a partial derivative, select a "variable of interest" and temporarily treat all other variables as constants. This allows you to examine the influence of just one variable at a time.
In the original exercise, finding \( g_x \) and \( g_y \) (the partial derivatives of \( g \) with respect to \( x \) and \( y \) respectively) required us to treat \( y \) as a constant for \( g_x \) and \( x \) as a constant for \( g_y \).
The results provided insight into how changes in one variable affect the function, holding the others steady. This is particularly useful when you need to determine sensitivity or rate of change in one specific direction within a multidimensional space.
Calculus Steps
Solving calculus problems, especially in multivariable calculus, involves clear methodical steps. Focusing on organized steps is key to mastering these problems.
1. **Identify the function and variables**: Determine which variables the function depends on. Understand the relationship between them and the function itself.
2. **Choose the variable for differentiation**: When finding a partial derivative, decide which variable you're focusing on, treating others as constants temporarily.
3. **Apply the chain rule if needed**: For composite functions, utilize the chain rule to correctly find the derivative of the function with respect to the chosen variable.
4. **Compute the derivative**: Carry out the differentiation process. Simplify the results as needed to get the partial derivative.
Following these steps ensures that you systematically approach calculus problems, minimize errors, and deepen your understanding of complex functions. These skills not only help in academics but also in real-world problem-solving scenarios.
1. **Identify the function and variables**: Determine which variables the function depends on. Understand the relationship between them and the function itself.
2. **Choose the variable for differentiation**: When finding a partial derivative, decide which variable you're focusing on, treating others as constants temporarily.
3. **Apply the chain rule if needed**: For composite functions, utilize the chain rule to correctly find the derivative of the function with respect to the chosen variable.
4. **Compute the derivative**: Carry out the differentiation process. Simplify the results as needed to get the partial derivative.
Following these steps ensures that you systematically approach calculus problems, minimize errors, and deepen your understanding of complex functions. These skills not only help in academics but also in real-world problem-solving scenarios.
