Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-20, verify that $$ \frac{\partial^{2} f}{\partial y \partial x}=\frac{\partial^{2} f}{\partial x \partial y} $$ \(f(x, y)=3 e^{2 x} \cos y\)

Step-by-Step Solution

Answer

The mixed partial derivatives are equal: \(-6e^{2x} \sin y\).

Step 1: First Partial Derivative with Respect to x

Start by finding the partial derivative of the function \(f(x, y) = 3e^{2x} \cos y\) with respect to \(x\). Treat \(y\) as a constant:\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left(3e^{2x} \cos y\right) = 3 \cdot \cos y \cdot \frac{\partial}{\partial x} (e^{2x}) = 3 \cdot \cos y \cdot (2e^{2x}) = 6e^{2x} \cos y \]

Step 2: Partial Derivative of Result with Respect to y

Next, find the partial derivative of the result from Step 1 with respect to \(y\). Treat \(x\) as a constant:\[ \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (6e^{2x} \cos y) = 6e^{2x} \cdot \frac{\partial}{\partial y} (\cos y) = 6e^{2x} (-\sin y) = -6e^{2x} \sin y \]

Step 3: First Partial Derivative with Respect to y

Now, find the partial derivative of the original function \(f(x, y) = 3e^{2x} \cos y\) with respect to \(y\). Treat \(x\) as a constant:\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(3e^{2x} \cos y\right) = 3e^{2x} \cdot \frac{\partial}{\partial y} (\cos y) = 3e^{2x} (-\sin y) = -3e^{2x} \sin y \]

Step 4: Partial Derivative of Result with Respect to x

Find the partial derivative of the result from Step 3 with respect to \(x\). Treat \(y\) as a constant:\[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (-3e^{2x} \sin y) = -3 \sin y \cdot \frac{\partial}{\partial x} (e^{2x}) = -3 \sin y \cdot 2e^{2x} = -6e^{2x} \sin y \]

Step 5: Conclusion

Both mixed partial derivatives are equal, confirming the symmetry of second partial derivatives (Clairaut's theorem).\[ \frac{\partial^2 f}{\partial y \partial x} = -6e^{2x} \sin y = \frac{\partial^2 f}{\partial x \partial y} \]

Key Concepts

Mixed Partial Derivatives

To grasp the concept of mixed partial derivatives, it's essential to first understand what a partial derivative is. In mathematics, a partial derivative focuses on how a function changes as one particular variable changes, while other variables remain constant.

For functions with more than one variable, like our example function, mixed partial derivatives come into play. A mixed partial derivative is taken with respect to two different variables.

For instance, with the function \( f(x, y) = 3e^{2x} \cos y \), we investigated two mixed partial derivatives:

• First, we differentiated with respect to \( x \) and then with respect to \( y \), denoted as \( \frac{\partial^2 f}{\partial y \partial x} \).
• Then, we differentiated with respect to \( y \) and followed with \( x \), denoted as \( \frac{\partial^2 f}{\partial x \partial y} \).

Evaluating these shows how the function responds when both \( x \) and \( y \) simultaneously vary.

Understanding mixed partial derivatives is crucial in multivariable calculus, greatly assisting in the analysis of how functions behave when the input variables are interdependent.

Clairaut's Theorem

Clairaut's Theorem simplifies the computation of mixed partial derivatives. It states that if the second partial derivatives of a function are continuous, then the order of differentiation does not matter.

Consider the function \( f(x, y) = 3e^{2x} \cos y \). Applying Clairaut's Theorem, we verified:

• \( \frac{\partial^2 f}{\partial y \partial x} \)
• \( \frac{\partial^2 f}{\partial x \partial y} \)

Both mixed derivatives yielded the same result, \(-6e^{2x} \sin y\), confirming the theorem's assertion.

Through this theorem, computational tasks become more efficient, as it is not necessary to verify the equality for functions that meet the continuous criterion.

It's a fantastic tool for confirming that our work with partial derivatives aligns consistently, ensuring that the symmetries in functions are used to simplify complex calculations.

Functions of Two Variables

Functions of two variables, like \( f(x, y) = 3e^{2x} \cos y \), extend calculus into the realm of surfaces and beyond linear graphs. Each point on such a surface corresponds to an \( (x, y) \) pair, essentially depicting a three-dimensional landscape.

Understanding these functions involves visualizing how the variables influence outcomes. These functions often model real-world scenarios, such as predicting temperatures across regions or analyzing economic output in various sectors.

For differentiation, functions of two variables require partial derivatives, which show the function's tilt or slope in the direction of one variable, holding the other constant. With mixed derivatives, you explore how changes in one variable can indirectly affect another.

By mastering functions of multiple variables, you gain the ability to describe complex systems and resolve intricate problems in engineering, physics, and economics.