Chapter 3: Problem 33
Evaluate the partial derivatives at point \(P(0,1)\). Given \(f(x, y, z)=x^{3} y z^{2}\), find \(\frac{\partial^{2} f}{\partial x \partial y}\) and \(f_{z}(1,1,1) .\)
Short Answer
Expert verified
\(\frac{\partial^{2} f}{\partial x\partial y}(0,1,z)=0\) and \(\frac{\partial f}{\partial z}(1,1,1)=2.\)
Step by step solution
01
Find the first partial derivative with respect to x
To find the first partial derivative with respect to \(x\), treat \(y\) and \(z\) as constants. Differentiate \(f(x, y, z)=x^{3} y z^{2}\) with respect to \(x\): \[ \frac{\partial f}{\partial x} = 3x^2 y z^2. \]
02
Differentiate the above result with respect to y
Taking the result from Step 1, differentiate \( 3x^2 y z^2 \) with respect to \(y,\) treating \(x\) and \(z\) as constants: \[ \frac{\partial^{2} f}{\partial y\partial x} = 3x^2 z^2.\]
03
Evaluate at the point P(0,1,z)
We don't need the complete point since \(z\) will still be a factor, so plug \(x=0\) and \(y=1\) into the equation from Step 2: \[ \frac{\partial^{2} f}{\partial y\partial x} (0,1,z) = 3(0)^2 z^2 = 0. \]
04
Verify that mixed partial derivative is continuous
Since \(f(x, y, z)\) is a polynomial function, the mixed partial derivatives should be continuous. Therefore, \[ \frac{\partial^{2} f}{\partial x\partial y} = \frac{\partial^{2} f}{\partial y\partial x}, \] which means \(\frac{\partial^{2} f}{\partial x\partial y}(0,1,z) = 0.\)
05
Find the partial derivative with respect to z
Differentiate \(f(x, y, z)=x^{3} y z^{2}\) with respect to \(z\), treating \(x\) and \(y\) as constants:\[ \frac{\partial f}{\partial z} = 2x^3 y z.\]
06
Evaluate \(\frac{\partial f}{\partial z}\) at \((1,1,1)\)
Substitute \(x = 1\), \(y = 1\), and \(z = 1\) into the previous result:\[ \frac{\partial f}{\partial z}(1,1,1) = 2(1)^3 (1)(1) = 2. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mixed Partial Derivatives
In calculus, the mixed partial derivatives provide insight into how a function changes with respect to two or more variables. When we talk about mixed partial derivatives, we're referring to the derivatives of a function taken first with respect to one variable and then with respect to another.
In the context of our exercise, consider the function \( f(x, y, z) = x^3 y z^2 \). We are interested in evaluating the mixed partial derivative \( \frac{\partial^2 f}{\partial x \partial y} \), also denoted as \( \frac{\partial^2 f}{\partial y \partial x} \).
By first differentiating with respect to \(x\), holding \(y\) and \(z\) constant, we obtain \( 3x^2 y z^2 \). Differentiating this result with respect to \(y\), while holding \(x\) and \(z\) constant, we arrive at \( 3x^2 z^2 \).
In the context of our exercise, consider the function \( f(x, y, z) = x^3 y z^2 \). We are interested in evaluating the mixed partial derivative \( \frac{\partial^2 f}{\partial x \partial y} \), also denoted as \( \frac{\partial^2 f}{\partial y \partial x} \).
By first differentiating with respect to \(x\), holding \(y\) and \(z\) constant, we obtain \( 3x^2 y z^2 \). Differentiating this result with respect to \(y\), while holding \(x\) and \(z\) constant, we arrive at \( 3x^2 z^2 \).
- This derivative indicates how the function \(f\) behaves as both \(x\) and \(y\) change while \(z\) remains fixed.
- As a polynomial function, \(f(x, y, z)\) ensures that the order of differentiation does not matter due to the Clairaut's Theorem, which means \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \).
Polynomial Functions
A polynomial function is a mathematical expression consisting of variables, coefficients, and exponents arranged in a specific manner. In our exercise, the function \( f(x, y, z) = x^3 y z^2 \) showcases these attributes with \(x, y,\) and \(z\) as variables, combined using multiplication and exponentiation.
These types of functions are incredibly important because they have properties that simplify calculus operations:
These types of functions are incredibly important because they have properties that simplify calculus operations:
- They are differentiable everywhere in the domain of real numbers, meaning derivatives can be found without encountering undefined expressions.
- Polynomials are continuous, making it easier to apply calculus theorems like Clairaut's on mixed partial derivatives.
- The power rule is often used to differentiate polynomial terms, allowing for straightforward calculations.
Derivation Steps
The process of finding derivatives, including mixed partial derivatives, involves a systematic approach where we treat certain variables as constants while differentiating with respect to others. This step-by-step progression is essential for tackling functions with multiple variables. Let's break down the process:
- **Step 1: Partial Derivative with Respect to a Single Variable** - Begin by selecting one variable to differentiate with respect to, treating all other variables as constants. For instance, we find \( \frac{\partial f}{\partial x} \) of \( f(x, y, z) = x^3 y z^2 \) by applying the power rule to obtain \( 3x^2 y z^2 \).
- **Step 2: Mixed Partial Derivative** - To find the next derivative with respect to another variable, differentiate the result of the first step, treating the initial variable as a constant. Here, doing this with respect to \(y\) gives \(3x^2 z^2\).
- **Step 3: Evaluate at Given Points or Limits** - Substitute specific values into the derived expressions to evaluate the derivative at particular points, like at \( P(0, 1, z) \) and \( (1, 1, 1) \).