Problem 4
Question
Find the four second partial derivatives of \(f(x, y)=3 x^{2} y+x y^{3}.\)
Step-by-Step Solution
Verified Answer
Answer: The four second partial derivatives are \(f_{xx} = 6y\), \(f_{xy} = 6x + 3y^2\), \(f_{yx} = 6x + 3y^2\), and \(f_{yy} = 6xy\).
1Step 1: Find the first partial derivatives
We have to find the first partial derivatives of the given function \(f(x, y)=3 x^{2} y+x y^{3}\). To find the partial derivatives, we will differentiate the function with respect to x and y.
- The first partial derivative with respect to x:
\(f_x = \frac{\partial f(x, y)}{\partial x} = \frac{\partial (3 x^{2} y+x y^{3})}{\partial x} = 6xy + y^3\)
- The first partial derivative with respect to y:
\(f_y = \frac{\partial f(x, y)}{\partial y} = \frac{\partial (3 x^{2} y+x y^{3})}{\partial y} = 3x^2 + 3xy^2\)
2Step 2: Find the second partial derivatives
Now, we will differentiate the first partial derivatives with respect to x and y.
1. The second partial derivative with respect to x (i.e., differentiate \(f_x\) with respect to x):
\(f_{xx} = \frac{\partial^2 f(x, y)}{\partial x^2} = \frac{\partial (6xy + y^3)}{\partial x} = 6y\)
2. The second partial derivative with respect to x and then y (i.e., differentiate \(f_x\) with respect to y):
\(f_{xy} = \frac{\partial^2 f(x, y)}{\partial x \partial y} = \frac{\partial (6xy + y^3)}{\partial y} = 6x + 3y^2\)
3. The second partial derivative with respect to y and then x (i.e., differentiate \(f_y\) with respect to x):
\(f_{yx} = \frac{\partial^2 f(x, y)}{\partial y \partial x} = \frac{\partial (3x^2 + 3xy^2)}{\partial x} = 6x + 3y^2\)
4. The second partial derivative with respect to y (i.e., differentiate \(f_y\) with respect to y):
\(f_{yy} = \frac{\partial^2 f(x, y)}{\partial y^2} = \frac{\partial (3x^2 + 3xy^2)}{\partial y} = 6xy\)
So, the four second partial derivatives are: \(f_{xx} = 6y\), \(f_{xy} = 6x + 3y^2\), \(f_{yx} = 6x + 3y^2\), and \(f_{yy} = 6xy\).
Key Concepts
Multivariable CalculusSecond Partial DerivativesDifferentiationFunctions of Multiple Variables
Multivariable Calculus
Multivariable calculus is the branch of mathematics that extends calculus to functions of more than one variable.
While single-variable calculus deals with functions of a single variable such as \(f(x)\), multivariable calculus involves functions of two or more variables like \(f(x, y)\).
This expands the capabilities of calculus from dealing with lines and curves to dealing with surfaces and higher-dimensional analogues.
While single-variable calculus deals with functions of a single variable such as \(f(x)\), multivariable calculus involves functions of two or more variables like \(f(x, y)\).
This expands the capabilities of calculus from dealing with lines and curves to dealing with surfaces and higher-dimensional analogues.
- It's crucial for applications involving changes across multiple dimensions, such as in physics to describe the world around us.
- In the context of optimization or economics, understanding how a function behaves across multiple variables allows us to solve real-world problems.
Second Partial Derivatives
Second partial derivatives give insights into the curvature and concavity of surfaces formed by functions of multiple variables.
Just as a second derivative in single-variable calculus expresses the rate of change of the rate of change, second partial derivatives extend this concept to multiple dimensions.
Just as a second derivative in single-variable calculus expresses the rate of change of the rate of change, second partial derivatives extend this concept to multiple dimensions.
- They are found by differentiating the first partial derivatives.
- For example, in our function \(f(x, y)\), the second partial derivatives include \(f_{xx}\), \(f_{yy}\), \(f_{xy}\), and \(f_{yx}\).
- \(f_{xx}\) corresponds to concavity in the x direction.
- \(f_{yy}\) indicates concavity in the y direction.
- \(f_{xy}\) and \(f_{yx}\), often equal across many functions, describe the mixed rate of change respecting both x and y.
Differentiation
Differentiation is the process of finding the rate at which a function is changing at any given point.
In multivariable calculus, differentiation applies to functions with multiple variables by treating each variable separately while treating others as constants.
This leads to the concept of partial derivatives, expressed as \(\frac{\partial}{\partial x}\) or \(\frac{\partial}{\partial y}\).
In multivariable calculus, differentiation applies to functions with multiple variables by treating each variable separately while treating others as constants.
This leads to the concept of partial derivatives, expressed as \(\frac{\partial}{\partial x}\) or \(\frac{\partial}{\partial y}\).
- Partial derivatives, like \(f_x\) or \(f_y\), denote sensitivity of the function \(f(x, y)\) to changes in one specific variable.
- They are foundational for defining directions of greatest increase or decrease across surfaces.
Functions of Multiple Variables
Functions of multiple variables are a step beyond the basic functions that depend only on a single input. These functions, like \(f(x, y)\), rely on more than one independent variable to produce a result.
They allow for complex modeling of systems where outcomes depend on several different factors.
Understanding how these variables interact through differentiation, including partial derivatives, is what makes solving real-world problems with multi-variable functions both challenging and broadly applicable.
They allow for complex modeling of systems where outcomes depend on several different factors.
- Commonly applied in fields such as physics, engineering, and economic modeling.
- Analyzing these functions involves examining different dimensions of change - not just linear.
Understanding how these variables interact through differentiation, including partial derivatives, is what makes solving real-world problems with multi-variable functions both challenging and broadly applicable.
Other exercises in this chapter
Problem 4
Let \(z=f(x, y), x=g(s, t),\) and \(y=h(s, t) .\) Explain how to find \(\partial z / \partial t\).
View solution Problem 4
What is the domain of \(h(x, y)=\sqrt{x-y} ?\)
View solution Problem 5
Lagrange multipliers in two variables Use Lagrange multipliers to find the maximum and minimum values of \(f\) (when they exist) subject to the given constraint
View solution Problem 5
What is the discriminant and how do you compute it?
View solution