Chapter 4: Problem 138
Given \(z=e^{x} \tan y,\) find \(\frac{\partial^{2} z}{\partial x \partial y}\) and \(\frac{\partial^{2} z}{\partial y \partial x}\)
Short Answer
Expert verified
Both mixed partial derivatives are equal: \(e^x \sec^2 y\).
Step by step solution
01
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: \[ \frac{\partial z}{\partial x} = e^x \tan y \].
02
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: \[ \frac{\partial z}{\partial y} = e^x \sec^2 y \].
03
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: \[ \frac{\partial^2 z}{\partial x \partial y} = e^x \sec^2 y \].
04
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:\[ \frac{\partial^2 z}{\partial y \partial x} = e^x \sec^2 y \].
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
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.
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:
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.
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.
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.