Problem 16 - APEX Calculus

Question

Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=(x+y)^{3} $$

Step-by-Step Solution

Answer

All second partial derivatives and mixed partial derivatives are the same: 6(x+y).

Step 1: Find the First Partial Derivative with respect to x

Differentiate the function \( f(x, y) = (x + y)^3 \) with respect to \( x \). Treat \( y \) as a constant. Using the chain rule, we get:\[ f_x = 3(x+y)^2 \cdot \frac{d}{dx}(x+y) = 3(x+y)^2 \cdot 1 = 3(x+y)^2. \]

Step 2: Find the First Partial Derivative with respect to y

Differentiate the function \( f(x, y) = (x + y)^3 \) with respect to \( y \). Treat \( x \) as a constant. Again, using the chain rule, we have:\[ f_y = 3(x+y)^2 \cdot \frac{d}{dy}(x+y) = 3(x+y)^2 \cdot 1 = 3(x+y)^2. \]

Step 3: Find the Second Partial Derivative with respect to x twice

Differentiate \( f_x = 3(x+y)^2 \) with respect to \( x \). We get:\[ f_{xx} = 6(x+y) \cdot \frac{d}{dx}(x+y) = 6(x+y) \cdot 1 = 6(x+y). \]

Step 4: Find the Second Partial Derivative with respect to y twice

Differentiate \( f_y = 3(x+y)^2 \) with respect to \( y \). We get:\[ f_{yy} = 6(x+y) \cdot \frac{d}{dy}(x+y) = 6(x+y) \cdot 1 = 6(x+y). \]

Step 5: Find the Second Mixed Partial Derivative \(f_{xy}\)

Differentiate \( f_x = 3(x+y)^2 \) with respect to \( y \). Thus:\[ f_{xy} = 6(x+y) \cdot \frac{d}{dy}(x+y) = 6(x+y) \cdot 1 = 6(x+y). \]

Step 6: Find the Second Mixed Partial Derivative \(f_{yx}\)

Differentiate \( f_y = 3(x+y)^2 \) with respect to \( x \). Thus:\[ f_{yx} = 6(x+y) \cdot \frac{d}{dx}(x+y) = 6(x+y) \cdot 1 = 6(x+y). \]

Key Concepts

Chain Rule

The chain rule is a key concept in calculus, especially useful when dealing with functions of multiple variables. It helps us find the derivative of a composite function by breaking it down into simpler parts.
In the context of the given exercise, we have the function \( f(x, y) = (x + y)^3 \), a composition of the outer function \( g(u) = u^3 \) and the inner function \( u = x+y \).
To find the partial derivative with respect to \( x \), we first differentiate the outer function: \( g'(u) = 3u^2 \). Then, we multiply by the derivative of the inner function with respect to \( x \), that’s simply \( \frac{d}{dx}(x+y) = 1 \). This results in:

• \( f_x = 3(x+y)^2 \).
• Similarly, for \( y \), \( f_y = 3(x+y)^2 \).

This application of the chain rule simplifies the process, making these calculations manageable, even with complex expressions.

Mixed Partial Derivatives

When we talk about mixed partial derivatives, we are referring to the order in which we differentiate a multivariable function with respect to its variables. Specifically, mixed derivatives involve differentiation in different orders.
In this exercise, the task is to compute \( f_{xy} \) and \( f_{yx} \).

• For \( f_{xy} \), we take the derivative of \( f_x = 3(x+y)^2 \) with respect to \( y \).
• Similarly, for \( f_{yx} \), we differentiate \( f_y = 3(x+y)^2 \) with respect to \( x \).

In both cases, we are essentially verifying that \( \frac{\partial^2 f}{\partial x \partial y} \) equals \( \frac{\partial^2 f}{\partial y \partial x} \). Here, the result is \( 6(x+y) \) for both, illustrating that mixed partial derivatives are usually equal (as they are in this example). This property is known as Clairaut's theorem, assuming continuity of the second derivatives.

Second Partial Derivatives

Second partial derivatives are crucial for understanding the behavior and curvature of multivariable functions. They involve taking the derivative of a function twice, with respect to one or more variables.
In our problem, after finding the first partial derivatives, we compute second partial derivatives like \( f_{xx} \) and \( f_{yy} \):

• For \( f_{xx} \), differentiate \( f_x = 3(x+y)^2 \) with respect to \( x \) again, which gives: \( f_{xx} = 6(x+y) \).
• For \( f_{yy} \), differentiate \( f_y = 3(x+y)^2 \) with respect to \( y \) which also yields \( f_{yy} = 6(x+y) \).

These second-order derivatives not only provide insights into the slope changes concerning each variable but also help in further analysis, such as finding critical points and analyzing the concavity of the function through the Hessian matrix.