Problem 30
Question
Find the four second-order partial derivatives. $$f(x, y)=3 x^{2} y-2 x y+4 y$$
Step-by-Step Solution
Verified Answer
The second-order partial derivatives are \( \frac{\partial^{2}f}{\partial x^{2}}=6y \), \( \frac{\partial^{2}f}{\partial y \partial x}=6x-2 \), \( \frac{\partial^{2}f}{\partial x \partial y}=6x-2 \), and \( \frac{\partial^{2}f}{\partial y^{2}}=0 \).
1Step 1: Derivative with respect to x
Start by calculating the first partial derivative of the function with respect to \( x \). The derivative of \( f(x, y) = 3x^2y - 2xy + 4y \) with respect to \( x \) is \( f_x(x, y) = \frac{\partial}{\partial x}(3x^2y - 2xy + 4y) = 6xy - 2y \).
2Step 2: Derivative of fx with respect to x
Now calculate the second partial derivative of \( f \) with respect to \( x \) again. We start from \( f_x(x, y) = 6xy - 2y \), then \( \frac{\partial^{2}f}{\partial x^{2}} = 6y \).
3Step 3: Derivative of fx with respect to y
Next, calculate the second partial derivative of \( f \) with respect to \( y \). Differentiate \( f_x(x, y) = 6xy - 2y \) with respect to \( y \) to get \( \frac{\partial^{2}f}{\partial y \partial x} = 6x - 2 \).
4Step 4: Derivative with respect to y
Find the first partial derivative of the function with respect to \( y \). Differentiate \( f(x, y) = 3x^2y - 2xy + 4y \) to get \( f_y(x, y) = \frac{\partial}{\partial y}(3x^2y - 2xy + 4y) = 3x^2 - 2x + 4 \).
5Step 5: Derivative of fy with respect to x
Calculate the second partial derivative of \( f \) with respect to \( x \). Start with \( f_y(x, y) = 3x^2 - 2x + 4 \) and differentiate with respect to \( x \), \( \frac{\partial^{2}f}{\partial x \partial y} = 6x - 2 \).
6Step 6: Derivative of fy with respect to y
Lastly, calculate the second partial derivative of \( f \) with respect to \( y \) again. Starting from \( f_y(x, y) = 3x^2 - 2x + 4 \) yields \( \frac{\partial^{2}f}{\partial y^{2}} = 0 \), since the derivative of a constant is zero.
Key Concepts
Partial DifferentiationMultivariable CalculusDerivatives of Functions of Two Variables
Partial Differentiation
Partial differentiation is a central concept in multivariable calculus, mainly used when we have a function with more than one variable. In this case, we differentiate the function with respect to one variable while keeping the other variables constant. Let's look at it through an example involving the function:
- The given function: \( f(x, y) = 3x^2y - 2xy + 4y \)
- \( f_x(x, y) = \frac{\partial}{\partial x}(3x^2y - 2xy + 4y) = 6xy - 2y \).
- \( f_y(x, y) = \frac{\partial}{\partial y}(3x^2y - 2xy + 4y) = 3x^2 - 2x + 4 \).
Multivariable Calculus
Multivariable calculus extends the basic concepts of calculus, like differentiation and integration, to functions of several variables. In the context of our problem, the function \( f(x, y) \) is a real-valued function of two variables, \( x \) and \( y \). Here are some of the basic ideas:
- A function of two variables like \( f(x, y) \) maps ordered pairs of real numbers to real numbers.
- To approach this, we use concepts such as partial derivatives to find the rates of change of the function with respect to each individual variable.
- Beyond basic differentiation, multivariable calculus also explores gradients, directional derivatives, and multiple integrals.
Derivatives of Functions of Two Variables
In multivariable functions, derivatives describe how a function changes at a certain point with respect to changes in each variable. Consider the second-order partial derivatives which are especially informative for understanding the concavity and convexity of functions:
- First-order derivatives like \( f_x \) and \( f_y \) reveal the slope or rate of change in the \( x \) and \( y \) directions respectively.
- Second-order partial derivatives give us additional insights:
- \( \frac{\partial^2 f}{\partial x^2} = 6y \): Change in rate along the \( x \)-direction.
- \( \frac{\partial^2 f}{\partial y \partial x} = 6x - 2 \): Change in the slope of \( f_x \) along \( y \)-direction.
- \( \frac{\partial^2 f}{\partial x \partial y} = 6x - 2 \): Same as the previous because mixed derivatives are equal under Schwartz's theorem.
- \( \frac{\partial^2 f}{\partial y^2} = 0 \): Indicates no curvature in the \( y \)-direction.
Other exercises in this chapter
Problem 29
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