Chapter 14: Problem 44
Find all the second-order partial derivatives of the functions. \(h(x, y)=x e^{y}+y+1\)
Short Answer
Expert verified
\( \frac{\partial^2 h}{\partial x^2} = 0, \frac{\partial^2 h}{\partial y^2} = x e^y, \frac{\partial^2 h}{\partial x \partial y} = e^y, \frac{\partial^2 h}{\partial y \partial x} = e^y.\)
Step by step solution
01
Find the First Partial Derivative with Respect to x
To find the first partial derivative of the function \(h(x, y)=x e^{y}+y+1\) with respect to \(x\), treat \(y\) as a constant. The derivative of \(x e^y\) with respect to \(x\) is \(e^y\), and the derivative of \(y+1\) is 0, so: \(\frac{\partial h}{\partial x} = e^y.\)
02
Find the First Partial Derivative with Respect to y
Now find the first partial derivative of \(h(x, y)=x e^{y}+y+1\) with respect to \(y\), treating \(x\) as a constant. The derivative of \(x e^y\) with respect to \(y\) is \(x e^y\), the derivative of \(y\) is 1, so: \(\frac{\partial h}{\partial y} = x e^y + 1.\)
03
Find the Second Partial Derivative with Respect to x Twice
Differentiate \(\frac{\partial h}{\partial x} = e^y\) with respect to \(x\) again. Since \(e^y\) is independent of \(x\), this derivative is 0: \(\frac{\partial^2 h}{\partial x^2} = 0.\)
04
Find the Second Partial Derivative with Respect to y Twice
Differentiate \(\frac{\partial h}{\partial y} = x e^y + 1\) with respect to \(y\) again. The derivative of \(x e^y\) with respect to \(y\) is \(x e^y\), and the derivative of 1 is 0: \(\frac{\partial^2 h}{\partial y^2} = x e^y.\)
05
Find the Mixed Second Partial Derivative \( \frac{ \partial^2 h}{\partial x \partial y} \)
To find \( \frac{ \partial^2 h}{\partial x \partial y} \), differentiate \( \frac{\partial h}{\partial x} = e^y \) with respect to \(y\). The derivative of \(e^y\) with respect to \(y\) is \(e^y\): \(\frac{\partial^2 h}{\partial x \partial y} = e^y.\)
06
Find the Mixed Second Partial Derivative \( \frac{ \partial^2 h}{\partial y \partial x} \)
To find \( \frac{ \partial^2 h}{\partial y \partial x} \), differentiate \( \frac{\partial h}{\partial y} = x e^y + 1 \) with respect to \(x\). The derivative of \(x e^y\) with respect to \(x\) is \(e^y\), and the derivative of 1 with respect to \(x\) is 0: \(\frac{\partial^2 h}{\partial y \partial x} = e^y.\)
07
Conclusion: List of All Second-Order Partial Derivatives
The second-order partial derivatives are:\[\frac{\partial^2 h}{\partial x^2} = 0,\]\[\frac{\partial^2 h}{\partial y^2} = x e^y,\]\[\frac{\partial^2 h}{\partial x \partial y} = e^y,\]\[\frac{\partial^2 h}{\partial y \partial x} = e^y.\] Notice that the mixed partial derivatives are equal as expected.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second Order Derivatives
Second order derivatives are essentially the derivatives of derivatives. For a function of two variables such as \( h(x, y) = x e^y + y + 1 \), the process involves differentiating the first order derivatives again with respect to the variables \( x \) and \( y \). The second order partial derivatives provide insights into the function's curvature along different axes.
Let's break it down:
Let's break it down:
- The second order partial derivative with respect to \( x \), \( \frac{\partial^2 h}{\partial x^2} \), involves differentiating the first derivative \( \frac{\partial h}{\partial x} = e^y \) again with respect to \( x \). Since \( e^y \) is independent of \( x \), the result is 0.
- The second order partial derivative with respect to \( y \), \( \frac{\partial^2 h}{\partial y^2} \), requires differentiating \( \frac{\partial h}{\partial y} = x e^y + 1 \) again with respect to \( y \). Here, the derivative of \( x e^y \) is \( x e^y \), while 1 is a constant with a derivative of 0, resulting in \( x e^y \).
Mixed Partial Derivatives
Mixed partial derivatives occur when we take partial derivatives of a function successively with respect to different variables. For functions like \( h(x, y) = x e^y + y + 1 \), these derivatives can tell us about how the function changes when both variables are altered together.
In this exercise:
In this exercise:
- The mixed partial derivative \( \frac{\partial^2 h}{\partial x \partial y} \) is found by differentiating \( \frac{\partial h}{\partial x} = e^y \) with respect to \( y \). This results in \( e^y \) as the derivative of \( e^y \) with respect to \( y \) is \( e^y \).
- The reverse order, \( \frac{\partial^2 h}{\partial y \partial x} \), involves differentiating \( \frac{\partial h}{\partial y} = x e^y + 1 \) with respect to \( x \), which gives \( e^y \) as \( e^y \) is the coefficient of \( x \) in the term \( x e^y \).
Partial Derivatives of Functions
Partial derivatives are fundamental in dealing with functions of multiple variables because they allow us to see how a function changes as one specific variable changes, while keeping others constant. Taking a closer look at the function \( h(x, y) = x e^y + y + 1 \), we focus on how each term reacts when isolated variable changes are made.
Here's the process for finding the first order partial derivatives:
Here's the process for finding the first order partial derivatives:
- For \( \frac{\partial h}{\partial x} \), consider \( y \) constant. The derivative of \( x e^y \) with respect to \( x \) is \( e^y \), while \( y + 1 \) gives 0, making \( \frac{\partial h}{\partial x} = e^y \).
- For \( \frac{\partial h}{\partial y} \), fix \( x \) and differentiate. \( x e^y \) with respect to \( y \) becomes \( x e^y \) and \( y + 1 \)'s derivative is simply 1, resulting in \( \frac{\partial h}{\partial y} = x e^y + 1 \).