Problem 36 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the second-order partial derivatives of the function. In each case, show that the mixed partial derivatives \(f_{x y}\) and \(f_{y x}\) are equal. \(f(x, y)=x^{3}+x^{2} y+x+4\)

Step-by-Step Solution

Answer

The second-order partial derivatives of the function \(f(x, y) = x^3 + x^2 y + x + 4\) are as follows: 1. \(f_{xx}(x, y) = 6x + 2y\) 2. \(f_{yy}(x, y) = 0\) 3. \(f_{xy}(x, y) = 2x\) 4. \(f_{yx}(x, y) = 2x\) We can verify that the mixed partial derivatives, \(f_{xy}(x, y)\) and \(f_{yx}(x, y)\), are equal, as both are equal to \(2x\).

Step 1: Calculate the first partial derivatives of \(f(x, y)\) with respect to x and y

To find the second-order partial derivatives, first, we need to find the first-order partial derivatives of the function with respect to both x and y. We can do this by applying the rules for partial differentiation. The first-order partial derivative with respect to x: \[f_x(x, y) = \frac{\partial}{\partial x}(x^3 + x^2y + x + 4) = 3x^2 + 2xy + 1\] The first-order partial derivative with respect to y: \[f_y(x, y) = \frac{\partial}{\partial y}(x^3 + x^2y + x + 4) = x^2\]

Step 2: Calculate the second-order partial derivatives

Now, we will differentiate the first-order partial derivatives obtained in step 1 to find the second-order partial derivatives. The second partial derivative with respect to x: \[f_{xx}(x, y) = \frac{\partial^2}{\partial x^2}(x^3 + x^2y + x + 4) = \frac{\partial}{\partial x}(3x^2 + 2xy + 1) = 6x + 2y\] The second partial derivative with respect to y: \[f_{yy}(x, y) = \frac{\partial^2}{\partial y^2}(x^3 + x^2y + x + 4) = \frac{\partial}{\partial y}(x^2) = 0\] The mixed partial derivative first with respect to x and then to y: \[f_{xy}(x, y) = \frac{\partial^2}{\partial x \partial y}(x^3 + x^2y + x + 4) = \frac{\partial}{\partial x}(x^2) = 2x\] The mixed partial derivative first with respect to y and then to x: \[f_{yx}(x, y) = \frac{\partial^2}{\partial y \partial x}(x^3 + x^2y + x + 4) = \frac{\partial}{\partial y}(3x^2 + 2xy + 1) = 2x\]

Step 3: Verify that the mixed partial derivatives are equal

Finally, we need to verify that \(f_{xy}(x, y) = f_{yx}(x, y)\). We obtained \(f_{xy}(x, y) = 2x\) and \(f_{yx}(x, y) = 2x\). Both mixed partial derivatives are the same, so we have verified that they are equal.

Key Concepts

Second-Order Partial Derivatives

Second-order partial derivatives are crucial when analyzing the behavior and curvature of multivariable functions. When you find these derivatives, you are essentially taking the partial derivative of a partial derivative.
Let's consider our given function:

• First, we differentiate with respect to one variable, such as finding \( f_x(x, y) = 3x^2 + 2xy + 1 \),
• Then, we differentiate \( f_x \) again with respect to the same variable to get \( f_{xx}(x, y) = 6x + 2y \).

Doing this process with both variables will yield all possible second-order combinations. Each outcome provides insights into how one variable impacts the function as changes occur in itself and others. Understanding these concavities or convexities in functions allows predictions about the function's shape and behavior.

Mixed Partial Derivatives

Mixed partial derivatives involve differentiating twice, but each time with a different variable. For example, with our function, when we compute

• First with respect to \( x \) and then \( y \), we find \( f_{xy}(x, y) = 2x \).
• Conversely, differentiating first with \( y \) and then \( x \) yields \( f_{yx}(x, y) = 2x \).

A noteworthy property of mixed partial derivatives comes from Clairaut's theorem, which states that under certain conditions, these mixed derivatives are equal: \( f_{xy} = f_{yx} \).
This theorem assures us the path we take – whether \( x \) to \( y \) or \( y \) to \( x \) – results in the same rate of change, given the function is well-behaved. Understanding and verifying this helps in ensuring accuracy and consistency in multivariable calculus computations.

Partial Differentiation

Partial differentiation is the process of differentiating a function of multiple variables with respect to only one variable at a time, treating all other variables as constants.
For instance, let's take the function \( f(x, y)=x^{3}+x^{2}y+x+4 \):

• To find the partial derivative with respect to \( x \), noted as \( f_x \), you'd treat \( y \) as a constant, focusing only on changes in \( x \).
• Similarly, to find \( f_y \), you'd differentiate with respect to \( y \) while treating \( x \) as a constant.

This method allows us to isolate the effect of changes in one variable at a time, a vital technique when dealing with complex functions where variables impact different dimensions.
Being comfortable with partial differentiation lays the groundwork for exploring more advanced topics like optimization and differential equations.