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In Exercises \(51-54,\) verify that \(w_{x y}=w_{y x}\) $$w=\ln (2 x+3 y)$$

Short Answer

Expert verified
\(w_{xy} = w_{yx}\); both are \(-\frac{6}{(2x + 3y)^2}\).

Step by step solution

01

Find the Partial Derivative with respect to x

Given the function \(w = \ln(2x + 3y)\), we need to find the partial derivative with respect to \(x\). The derivative of \(\ln u\) is \(\frac{1}{u}\), and by chain rule, the derivative of \(2x + 3y\) with respect to \(x\) is 2. Thus, the partial derivative is \(w_x = \frac{d}{dx}[\ln(2x + 3y)] = \frac{1}{2x + 3y} \cdot 2 = \frac{2}{2x + 3y}\).
02

Differentiate w_x with respect to y

Now we differentiate \(w_x = \frac{2}{2x + 3y}\) with respect to \(y\). Use the chain rule for this step. The derivative of \(\frac{1}{u}\) is \(-\frac{1}{u^2}\), and the derivative of \(2x + 3y\) with respect to \(y\) is 3. Thus, \(w_{xy} = \frac{d}{dy}\left(\frac{2}{2x + 3y}\right) = -\frac{2 \times 3}{(2x + 3y)^2} = -\frac{6}{(2x + 3y)^2}\).
03

Find the Partial Derivative with respect to y

Next, find the partial derivative of \(w\) with respect to \(y\). Using the chain rule again, \(w_y = \frac{1}{2x + 3y} \cdot 3 = \frac{3}{2x + 3y}\).
04

Differentiate w_y with respect to x

Differentiate \(w_y = \frac{3}{2x + 3y}\) with respect to \(x\). Again, use the chain rule. The derivative of \(\frac{1}{u}\) is \(-\frac{1}{u^2}\), and the derivative of \(2x + 3y\) with respect to \(x\) is 2. Thus, \(w_{yx} = \frac{d}{dx}\left(\frac{3}{2x + 3y}\right) = -\frac{3 \times 2}{(2x + 3y)^2} = -\frac{6}{(2x + 3y)^2}\).
05

Verify the equality w_{xy} = w_{yx}

Compare \(w_{xy}\) and \(w_{yx}\). From Step 2, we have \(w_{xy} = -\frac{6}{(2x + 3y)^2}\). From Step 4, we have \(w_{yx} = -\frac{6}{(2x + 3y)^2}\). Thus, \(w_{xy} = w_{yx}\), verifying the equality.

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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 significant concept when working with derivatives, especially when dealing with composite functions. In mathematics, the chain rule helps us differentiate a function that is nested within another function. This is common when working with functions of multiple variables, where the inner function is often dependent on other variables.

For example, in the function given by the exercise \(w = \ln(2x + 3y)\), the expression \(2x + 3y\) can be viewed as an inner function. When taking the derivative of \(w\) with respect to \(x\) or \(y\), the chain rule tells us to:
  • First take the derivative of the outer function assuming the inner function is unchanged.
  • Then multiply it by the derivative of the inner function with respect to the variable of interest.
This method allows us to accurately handle the changes in the nested parts of the function. Using the chain rule, we found that:
  • With respect to \(x\): \(w_x = \frac{2}{2x + 3y}\)
  • With respect to \(y\): \(w_y = \frac{3}{2x + 3y}\)
The chain rule is critical for determining how composite functions behave under differentiation.
Functions of Multiple Variables
Functions of multiple variables are functions that depend on more than one input. In calculus, these functions are often expressed as \(f(x, y, z)\) to denote a function depending on variables \(x, y, ext{ and } z\). Understanding these functions require us to look at how changes in one variable affect the function while keeping others constant.

For example, in the function \(w = \ln(2x + 3y)\), \(w\) depends on both \(x\) and \(y\). This means we can consider how \(w\) changes with respect to \(x\) alone, finding the partial derivative \(w_x\), while treating \(y\) as a constant, and vice-versa for \(w_y\).
  • Partial derivatives indicate the rate of change of the function with respect to one variable at a time.
  • This is crucial in fields like engineering and physics, where multiple factors often change simultaneously.
By understanding the behavior of multivariable functions, we can model complex systems more accurately.
Second Derivatives
Second derivatives provide insights into the curvature or concavity of a function, especially in functions with multiple variables. They represent the derivative of the derivative, giving us another layer of information about how a function behaves.

In our exercise, we explored second partial derivatives. After finding the first partial derivatives \(w_x\) and \(w_y\), we further differentiated:
  • First, \(w_x\) with respect to \(y\) to find \(w_{xy}\).
  • Then, \(w_y\) with respect to \(x\) to find \(w_{yx}\).
Second derivatives are significant in verifying equality like \(w_{xy} = w_{yx}\), a fundamental property of continuous functions with continuous second derivatives, known as Schwarz's theorem.
  • This verification demonstrates consistency and continuity in the behavior of the function across its variables.
Understanding second derivatives helps in determining how quickly changes in the function accelerate or decelerate, which is instrumental in predicting the behavior of dynamic systems.

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