Chapter 14: Problem 62
Verify that the conclusion of Clairaut's Theorem holds, that is, \(u_{x y}=u_{y x}\) $$ u=\ln (x+2 y) $$
Short Answer
Expert verified
The mixed partial derivatives \(u_{xy}\) and \(u_{yx}\) are equal, verifying Clairaut's Theorem.
Step by step solution
01
Compute First Partial Derivatives
First, let's compute the partial derivatives of the function \(u\). Start by finding \(u_x\) and \(u_y\). The function given is \(u = \ln(x + 2y)\).**Partial derivative with respect to \(x\):**\[ u_x = \frac{\partial}{\partial x} \ln(x + 2y) = \frac{1}{x + 2y} \times 1 = \frac{1}{x + 2y}\]**Partial derivative with respect to \(y\):**\[ u_y = \frac{\partial}{\partial y} \ln(x + 2y) = \frac{1}{x + 2y} \times 2 = \frac{2}{x + 2y}\]
02
Compute Second Partial Derivative \(u_{xy}\)
Now, compute the second partial derivative \(u_{xy}\) by differentiating \(u_x\) with respect to \(y\).\[ u_{xy} = \frac{\partial}{\partial y} \left(\frac{1}{x + 2y}\right) = \frac{-1\times 2}{(x + 2y)^2} = \frac{-2}{(x + 2y)^2}\]
03
Compute Second Partial Derivative \(u_{yx}\)
Next, compute the second partial derivative \(u_{yx}\) by differentiating \(u_y\) with respect to \(x\).\[ u_{yx} = \frac{\partial}{\partial x} \left(\frac{2}{x + 2y}\right) = \frac{-2\times 1}{(x + 2y)^2} = \frac{-2}{(x + 2y)^2}\]
04
Verify Clairaut's Theorem Conclusion
The goal is to verify that these mixed partial derivatives are equal, \(u_{xy} = u_{yx}\).Since \(u_{xy} = \frac{-2}{(x + 2y)^2}\) and \(u_{yx} = \frac{-2}{(x + 2y)^2}\), this shows:\[ u_{xy} = u_{yx}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
A partial derivative represents how a multivariable function changes as one variable changes while keeping the other variables constant. When dealing with a function like \( u = \ln(x + 2y) \), partial derivatives allow us to see how \( u \) changes as we modify \( x \) or \( y \) separately.
- For \( u_x \), we differentiate with respect to \( x \). This means treating \( y \) as a constant.
- For \( u_y \), we differentiate with respect to \( y \), keeping \( x \) fixed.
Mixed Partial Derivatives
Mixed partial derivatives involve taking two partial derivatives of a function, each with respect to different variables. The concept becomes particularly interesting because, for most well-behaved functions, mixed partial derivatives are equal, reflecting Clairaut's Theorem. Here's how it works:
- Start with the first partial derivative \( u_x \) and then take its derivative with respect to \( y \) to get \( u_{xy} \).
- Similarly, take the partial derivative \( u_y \) and differentiate it with respect to \( x \) to get \( u_{yx} \).
Logarithmic Functions
Logarithmic functions are an important component in mathematics, especially when connected with calculus concepts like derivatives. When you see a function like \( u = \ln(x + 2y) \), it highlights the inverse relationship with exponential functions.A few key points about logarithmic functions:
- The derivative of \( \ln(x) \) is \( \frac{1}{x} \), which indicates how the function rate of change behaves.
- In multivariable calculus, this translates to taking partial derivatives when the argument of the logarithm involves more than one variable.