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Magazine Chapter 14: Problem 48
Find all the second-order partial derivatives of the functions in Exercises \(41-50 .\) $$w=y e^{x^{2}-y}$$
Short Answer
Expert verified
The second-order partial derivatives are \( \frac{\partial^2 w}{\partial x^2} = 2ye^{x^2-y}(2x + 2x^2) \), \( \frac{\partial^2 w}{\partial y^2} = -e^{x^2-y} \), and mixed partials \( \frac{\partial^2 w}{\partial x \partial y} = 2x(1-y)e^{x^2-y} \).
Step by step solution
01
Understanding the Partial Derivatives
Partial derivatives are taken with respect to one variable, keeping other variables constant. Here, we need to find the second-order partial derivatives of the function \( w = y e^{x^2 - y} \).
02
First Partial Derivatives
First, find the partial derivatives with respect to \( x \) and \( y \).For \( \frac{\partial w}{\partial x} \):- Use the chain rule: \( \frac{\partial w}{\partial x} = y \cdot e^{x^2 - y} \cdot 2x = 2xy e^{x^2-y} \).For \( \frac{\partial w}{\partial y} \):- Differentiate \( w \) directly with respect to \( y \): \( \frac{\partial w}{\partial y} = e^{x^2 - y} - y \cdot e^{x^2 - y} = e^{x^2 - y}(1-y) \).
03
Second Order Partial Derivative with Respect to x Twice
Take the derivative of \( 2xy e^{x^2-y} \) with respect to \( x \) again.- \( \frac{\partial^2 w}{\partial x^2} = \frac{\partial }{\partial x}(2xy e^{x^2-y}) = 2y \cdot e^{x^2-y} (2x) + 2xy \cdot e^{x^2-y} (2x) = 2y e^{x^2-y}(2x) + 4x^2 y e^{x^2-y} = 2y e^{x^2-y}(2x + 2x^2) \).
04
Second Order Partial Derivative with Respect to y Twice
Take the derivative of \( e^{x^2-y}(1-y) \) with respect to \( y \) again.- Utilize the product rule: \( \frac{\partial^2 w}{\partial y^2} = \frac{\partial }{\partial y}(e^{x^2-y}(1-y)) = -e^{x^2-y} \), as the differentiation predominantly affects the terms \( (1 - y) \) and the exponential derivative loops back with a negative sign.
05
Mixed Partial Derivatives
Find the mixed partial derivatives by differentiating \( \frac{\partial w}{\partial x} \) with respect to \( y \) and \( \frac{\partial w}{\partial y} \) with respect to \( x \):1. \( \frac{\partial^2 w}{\partial y \partial x} = \frac{\partial }{\partial y} (2xy e^{x^2-y}) = 2x(e^{x^2-y})\cdot(-1) = -2x e^{x^2-y} \).2. \( \frac{\partial^2 w}{\partial x \partial y} = \frac{\partial }{\partial x} (e^{x^2-y}(1-y)) = (2x) e^{x^2-y}(1-y) = 2x(1-y)e^{x^2-y} \).Note: Since mixed partial derivatives are typically equal if the function is well-behaved, you can also verify these results.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Understanding partial derivatives is essential when dealing with functions of multiple variables. In these functions, we hold all variables except one constant while differentiating with respect to the chosen variable.
For example, when considering a function like \(w = y e^{x^2 - y}\), we compute the partial derivative with respect to \(x\) by treating \(y\) as a constant. Similarly, when taking a partial derivative with respect to \(y\), \(x\) is held constant.
For example, when considering a function like \(w = y e^{x^2 - y}\), we compute the partial derivative with respect to \(x\) by treating \(y\) as a constant. Similarly, when taking a partial derivative with respect to \(y\), \(x\) is held constant.
- Partial Derivative with respect to \(x\): It reveals how the function \(w\) changes as \(x\) varies while \(y\) remains the same.
- Partial Derivative with respect to \(y\): It shows how \(w\) changes with changes in \(y\) while \(x\) stays fixed.
Chain Rule
The chain rule is a crucial tool when dealing with compositions of functions, especially in multivariable calculus. It helps us differentiate functions within functions.
In our function \(w = y e^{x^2 - y}\), when finding the partial derivative \(\frac{\partial w}{\partial x}\), the chain rule comes into play because \(e^{x^2 - y}\) is a function composition.
By applying the chain rule, we differentiate the outer function while keeping the inner function intact, then multiply by the derivative of the inner function.
In our function \(w = y e^{x^2 - y}\), when finding the partial derivative \(\frac{\partial w}{\partial x}\), the chain rule comes into play because \(e^{x^2 - y}\) is a function composition.
By applying the chain rule, we differentiate the outer function while keeping the inner function intact, then multiply by the derivative of the inner function.
- In the term \(e^{x^2-y}, 2x\) represents the derivative of the inside function \(x^2-y\) with respect to \(x\).
- The resulting partial derivative simplifies to \(2xy e^{x^2-y}\).
Mixed Partial Derivatives
Mixed partial derivatives involve taking the derivative of a function with respect to two or more variables in succession.
This concept is central to understanding the behavior of multidimensional functions.
This is known as Clairaut's Theorem, and it provides symmetry in second derivatives.
This concept is central to understanding the behavior of multidimensional functions.
- Consider the function \(w = y e^{x^2 - y}\). To find mixed partial derivatives, differentiate twice: once with respect to \(x\) and once with respect to \(y\).
- For example, \(\frac{\partial^2 w}{\partial y \partial x} = -2x e^{x^2-y}\) and \(\frac{\partial^2 w}{\partial x \partial y} = 2x(1-y)e^{x^2-y}\).
This is known as Clairaut's Theorem, and it provides symmetry in second derivatives.
Product Rule
The product rule is vital when differentiating products of two or more functions. This rule helps calculate derivatives where the function involves multiplication between two distinct functions of the same or different variables.For the given function \(w = y e^{x^2 - y}\), when finding the second-order partial derivative with respect to \(y\) twice, we apply the product rule:
- The term \(e^{x^2-y}(1-y)\) involves a product of \(e^{x^2-y}\) and \((1-y)\).
- After applying the product rule, the derivative simplifies primarily because of the derivative impacting the \((1-y)\) term, leading to \(-e^{x^2-y}\).
