Problem 138 - Calculus 3

Question

Given \(z=e^{x} \tan y,\) find \(\frac{\partial^{2} z}{\partial x \partial y}\) and \(\frac{\partial^{2} z}{\partial y \partial x}\)

Step-by-Step Solution

Answer

Both mixed partial derivatives are equal: \(e^x \sec^2 y\).

Step 1: Differentiate with Respect to x

First, we need to differentiate the function \(z = e^x \tan y\) with respect to \(x\). Since \(\tan y\) is a constant with respect to \(x\), we get the partial derivative as: \[ \frac{\partial z}{\partial x} = e^x \tan y \].

Step 2: Differentiate with Respect to y

Now, let's differentiate the original function \(z = e^x \tan y\) with respect to \(y\). Here, \(e^x\) is treated as a constant and the derivative of \(\tan y\) with respect to \(y\) is \(\sec^2 y\). So, we have: \[ \frac{\partial z}{\partial y} = e^x \sec^2 y \].

Step 3: Find the Mixed Partial \(\frac{\partial^2 z}{\partial x \partial y}\)

Use the result from Step 1, \( \frac{\partial z}{\partial x} = e^x \tan y \), and differentiate it with respect to \(y\). The derivative of \(\tan y\) with respect to \(y\) is \(\sec^2 y\), so: \[ \frac{\partial^2 z}{\partial x \partial y} = e^x \sec^2 y \].

Step 4: Find the Mixed Partial \(\frac{\partial^2 z}{\partial y \partial x}\)

Now, use the result from Step 2, \( \frac{\partial z}{\partial y} = e^x \sec^2 y \), and differentiate it with respect to \(x\). Since \(\sec^2 y\) is constant with respect to \(x\), we simply have:\[ \frac{\partial^2 z}{\partial y \partial x} = e^x \sec^2 y \].

Step 5: Verify Equality of Mixed Partials

According to Clairaut's theorem, if the second mixed partial derivatives are continuous, they should be equal. Here, both \(\frac{\partial^2 z}{\partial x \partial y}\) and \(\frac{\partial^2 z}{\partial y \partial x}\) are equal to \(e^x \sec^2 y\), satisfying the theorem.

Key Concepts

Mixed Partial Derivatives

Mixed partial derivatives involve differentiating a function multiple times with respect to different variables. For a function, say\[ z = f(x, y), \]it involves finding derivatives like:\[ \frac{\partial^2 z}{\partial x \partial y} \quad \text{or} \quad \frac{\partial^2 z}{\partial y \partial x}. \]These derivatives measure how the function changes as two variables change simultaneously. In the example given, we calculated these mixed partial derivatives for the function \( z = e^x \tan y \).
To approach mixed partial derivatives:

• Differentiate with respect to one variable while treating other variables as constants.
• After finding the first partial derivative, differentiate again, this time with respect to the second variable.

The process shows how one variable affects the rate of change concerning the other. Checking: \( \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x} \),confirms interconnected rates of change.

Clairaut's Theorem

Clairaut's theorem is a fundamental principle in calculus related to partial derivatives. It states that if the mixed partial derivatives of a function are continuous around a point, they are equal. That is,\[ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x}. \]This condition ensures that the order of differentiation does not matter in a neighborhood of the point.
For Clairaut's theorem to hold:

• The function must have continuous second partial derivatives.
• The theorem often simplifies the computation and verification of partial derivatives.

In our example with the expression \( z = e^x \tan y \),we found:\[ \frac{\partial^2 z}{\partial x \partial y} = e^x \sec^2 y \quad \text{and} \quad \frac{\partial^2 z}{\partial y \partial x} = e^x \sec^2 y. \]This equality reaffirms Clairaut's theorem, showing the computations are correct and align with the theory.

Differentiation

Differentiation is the process of finding the derivative of a function. It's essential for understanding how a function changes at any given point. When dealing with functions of multiple variables such as \( z = f(x, y) \),special techniques are used, including partial derivatives.
Key points to remember about differentiation:

• It measures the rate of change of a function.
• For multiple variables, we use partial derivatives, treating other variables as constants.

In our problem, differentiation helps explore how changes in \( x \)and \( y \)affect \( z \).
We took partial derivatives:\[ \frac{\partial z}{\partial x} = e^x \tan y \quad \text{and} \quad \frac{\partial z}{\partial y} = e^x \sec^2 y, \]to explore these changes, leading to the computation of mixed partial derivatives. Differentiation helps analyze multi-variable functions' behavior, revealing key insights into their character.