Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

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: Calculate the First Partial Derivative with respect to x

To begin, we need to compute the first partial derivative of the function \(f(x, y) = 3e^{2x} \cos y\) with respect to \(x\). Differentiating \(3e^{2x}\) with respect to \(x\) gives \(6e^{2x}\), since \(\cos y\) is treated as a constant. Thus, \(\frac{\partial f}{\partial x} = 6e^{2x} \cos y\).

Step 2: Calculate the Second Order Partial Derivative \(\frac{\partial^2 f}{\partial y \partial x}\)

Next, we find the derivative of \(\frac{\partial f}{\partial x} = 6e^{2x} \cos y\) with respect to \(y\). The derivative of \(\cos y\) with respect to \(y\) is \(-\sin y\), so \(\frac{\partial^2 f}{\partial y \partial x} = 6e^{2x} (-\sin y) = -6e^{2x} \sin y\).

Step 3: Calculate the First Partial Derivative with respect to y

Now, compute the first partial derivative of the function \(f(x, y) = 3e^{2x} \cos y\) with respect to \(y\). Differentiating \(\cos y\) gives \(-\sin y\), so we have \(\frac{\partial f}{\partial y} = -3e^{2x} \sin y\).

Step 4: Calculate the Second Order Partial Derivative \(\frac{\partial^2 f}{\partial x \partial y}\)

Finally, differentiate \(\frac{\partial f}{\partial y} = -3e^{2x} \sin y\) with respect to \(x\). Since \(-3\sin y\) is constant with respect to \(x\), we have \(\frac{\partial^2 f}{\partial x \partial y} = (-3\sin y) (6e^{2x}) = -6e^{2x} \sin y\).

Step 5: Verify Equality of Mixed Second Order Derivatives

Both computed mixed partial derivatives are \(-6e^{2x} \sin y\). Therefore, \(\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}\), confirming the equality.

Key Concepts

Partial Differentiation

Partial differentiation is a technique in calculus used to find the rate of change of a function with respect to one variable, while keeping the other variables constant. This method allows us to understand how changes in one variable affect the function's output, without interference from changes in other variables.

For a function of two variables, such as \(f(x, y)\), we can compute the partial derivative with respect to \(x\) by treating \(y\) as a constant. Similarly, to find the partial derivative with respect to \(y\), \(x\) is held constant. This provides us with two first-order derivatives: \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

• Helps in determining how a small change in one variable affects the function.
• Essential for understanding multi-variable calculus problems.

Second Order Derivatives

Second order derivatives involve taking the derivative of a derivative, exploring the rate of change of the rate of change of a function. In the context of partial derivatives, the second order derivative tells us how the initial rate of change itself changes as the variables change.

You can compute mixed second order partial derivatives, such as \(\frac{\partial^2 f}{\partial x \partial y}\) and \(\frac{\partial^2 f}{\partial y \partial x}\), to see if changing \(x\) and then \(y\) has the same effect as changing \(y\) and then \(x\).

• Useful in understanding curvature and concavity in multivariable functions.
• Helps predict how the slope of change in one variable affects another variable.

Clairaut's Theorem

Clairaut’s theorem is foundational to the topic of mixed partial derivatives. It states that if the mixed partial derivatives are continuous near a point, then \(\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}\).

This theorem offers a very useful symmetry property for functions where mixed derivatives do not depend on the order of differentiation. Therefore, verifying mixed partial derivatives with Clairaut's theorem often confirms the function is smooth and well-behaved.

• Provides assurance that mixed partial derivatives can be interchanged.
• Ensures consistency and symmetry in second order derivatives.

Calculus

Calculus is the branch of mathematics that deals with continuous change and is divided into two main branches: differentiation and integration. Partial differentiation is an extension of single-variable differentiation and is critical in understanding changes in multivariable functions.

Calculus is used across various fields such as physics, engineering, economics, and anywhere changes and motion analysis are involved. It provides models to predict behaviors and find optimal solutions within certain conditions.

• Teaches how to calculate rates of change and understand functional behavior in changing systems.
• Lays the groundwork for further exploration in sciences and technology.