Problem 9
Question
Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=x^{2} y+3 x^{2}+4 y-5 $$
Step-by-Step Solution
Verified Answer
\( f_x = 2xy + 6x, f_y = x^2 + 4, f_{xx} = 2y + 6, f_{yy} = 0, f_{xy} = f_{yx} = 2x \)
1Step 1: Finding the Partial Derivative with respect to x, \(f_x\)
To find the partial derivative \(f_x\), we differentiate \(f(x, y) = x^2 y + 3x^2 + 4y - 5\) with respect to \(x\), treating \(y\) as a constant. \[ f_x = rac{ ext{d}}{ ext{d}x}(x^2y) + rac{ ext{d}}{ ext{d}x}(3x^2) + rac{ ext{d}}{ ext{d}x}(4y) + rac{ ext{d}}{ ext{d}x}(-5) \]\[ f_x = 2xy + 6x + 0 + 0 = 2xy + 6x \]
2Step 2: Finding the Partial Derivative with respect to y, \(f_y\)
To find the partial derivative \(f_y\), differentiate \(f(x, y)\) with respect to \(y\), treating \(x\) as a constant.\[ f_y = rac{ ext{d}}{ ext{d}y}(x^2 y) + rac{ ext{d}}{ ext{d}y}(3x^2) + rac{ ext{d}}{ ext{d}y}(4y) + rac{ ext{d}}{ ext{d}y}(-5) \]\[ f_y = x^2 + 0 + 4 + 0 = x^2 + 4 \]
3Step 3: Finding the Second Partial Derivative with respect to x, \(f_{xx}\)
To find \(f_{xx}\), differentiate \(f_x = 2xy + 6x\) with respect to \(x\).\[ f_{xx} = rac{ ext{d}}{ ext{d}x}(2xy) + rac{ ext{d}}{ ext{d}x}(6x) \]\[ f_{xx} = 2y + 6 \]
4Step 4: Finding the Second Partial Derivative with respect to y, \(f_{yy}\)
To find \(f_{yy}\), differentiate \(f_y = x^2 + 4\) with respect to \(y\).\[ f_{yy} = rac{ ext{d}}{ ext{d}y}(x^2) + rac{ ext{d}}{ ext{d}y}(4) \]\[ f_{yy} = 0 + 0 = 0 \]
5Step 5: Finding the Mixed Partial Derivative \(f_{xy}\)
To find \(f_{xy}\), differentiate \(f_x = 2xy + 6x\) with respect to \(y\).\[ f_{xy} = rac{ ext{d}}{ ext{d}y}(2xy) + rac{ ext{d}}{ ext{d}y}(6x) \]\[ f_{xy} = 2x + 0 = 2x \]
6Step 6: Finding the Mixed Partial Derivative \(f_{yx}\)
To find \(f_{yx}\), differentiate \(f_y = x^2 + 4\) with respect to \(x\).\[ f_{yx} = rac{ ext{d}}{ ext{d}x}(x^2) + rac{ ext{d}}{ ext{d}x}(4) \]\[ f_{yx} = 2x + 0 = 2x \]
Key Concepts
Multivariable CalculusSecond Partial DerivativesMixed Partial Derivatives
Multivariable Calculus
Multivariable calculus is a branch of calculus that deals with functions of more than one variable. In single-variable calculus, you work with functions like \(f(x)\), which depend on just one variable. However, in multivariable calculus, you encounter functions such as \(f(x, y)\), which depend on two or more variables. This makes understanding how changing one variable affects the function more complex and interesting.
For example, consider a function \(f(x, y) = x^2y + 3x^2 + 4y - 5\). Here, both \(x\) and \(y\) are independent variables. When dealing with such functions, you often need to compute derivatives concerning each variable separately, which we refer to as partial derivatives. These help in understanding how the function behaves as one variable changes while others are held constant. This becomes essential when analyzing and visualizing surfaces, as multivariable functions often represent 3-dimensional surfaces in space.
For example, consider a function \(f(x, y) = x^2y + 3x^2 + 4y - 5\). Here, both \(x\) and \(y\) are independent variables. When dealing with such functions, you often need to compute derivatives concerning each variable separately, which we refer to as partial derivatives. These help in understanding how the function behaves as one variable changes while others are held constant. This becomes essential when analyzing and visualizing surfaces, as multivariable functions often represent 3-dimensional surfaces in space.
Second Partial Derivatives
Second partial derivatives involve taking the partial derivative of a partial derivative. It's like differentiating a function twice, but for multivariable scenarios. After you find the first partial derivatives, such as \(f_x\) and \(f_y\) for a function \(f(x, y)\), you can differentiate them again with respect to the same variable to find the second partial derivatives.
In the earlier problem, we found \(f_x = 2xy + 6x\). To find the second partial derivative with respect to \(x\), denoted as \(f_{xx}\), we differentiate \(f_x\) with respect to \(x\) again to get \(f_{xx} = 2y + 6\). Similarly, if you have \(f_y = x^2 + 4\), differentiating again with respect to \(y\) gives \(f_{yy} = 0\). These second derivatives can help identify the curvature of a surface, determining whether points are maxima, minima, or saddle points.
In the earlier problem, we found \(f_x = 2xy + 6x\). To find the second partial derivative with respect to \(x\), denoted as \(f_{xx}\), we differentiate \(f_x\) with respect to \(x\) again to get \(f_{xx} = 2y + 6\). Similarly, if you have \(f_y = x^2 + 4\), differentiating again with respect to \(y\) gives \(f_{yy} = 0\). These second derivatives can help identify the curvature of a surface, determining whether points are maxima, minima, or saddle points.
Mixed Partial Derivatives
Mixed partial derivatives involve taking a partial derivative with respect to one variable and then taking the derivative of that result with respect to another variable. These are denoted as \(f_{xy}\) or \(f_{yx}\). The concept of mixed derivatives is crucial when analyzing functions for which multiple variables interact.
Using the earlier solution as an example, to find \(f_{xy}\), we start with the partial derivative \(f_x = 2xy + 6x\) and differentiate it with respect to \(y\), resulting in \(f_{xy} = 2x\). Conversely, by taking \(f_y = x^2 + 4\) and differentiating with respect to \(x\), we get \(f_{yx} = 2x\). Generally, for most functions that are continuous and differentiable, these mixed derivatives are equal; this equality is known as Clairaut's theorem. Understanding mixed partial derivatives is important in describing how changes in two variables simultaneously influence a function. They help in solving and analyzing equations in fields like physics and engineering, where such interactions are prevalent.
Using the earlier solution as an example, to find \(f_{xy}\), we start with the partial derivative \(f_x = 2xy + 6x\) and differentiate it with respect to \(y\), resulting in \(f_{xy} = 2x\). Conversely, by taking \(f_y = x^2 + 4\) and differentiating with respect to \(x\), we get \(f_{yx} = 2x\). Generally, for most functions that are continuous and differentiable, these mixed derivatives are equal; this equality is known as Clairaut's theorem. Understanding mixed partial derivatives is important in describing how changes in two variables simultaneously influence a function. They help in solving and analyzing equations in fields like physics and engineering, where such interactions are prevalent.
Other exercises in this chapter
Problem 9
In Exercises \(9-12,\) a function \(2=f(x, y)\) is given. Give the indicated approximation using the total differential. $$ \begin{array}{l} f(x, y)=\sqrt{x^{2}
View solution Problem 9
A set \(S\) is given. (a) Give one boundary point and one interior point, when possible, of \(S\). (b) State whether \(S\) is open, closed, or neither. (c) Stat
View solution Problem 9
Give the domain and range of the multivariable function. $$ f(x, y)=x-2 y $$
View solution Problem 10
Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, o
View solution