Problem 38 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The second partial derivatives are: \( f_{xx} = -e^{x}, f_{xy} = f_{yx} = 0, f_{yy} = 0. \)

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

To calculate the second order partial derivatives, we first find the first partial derivative of the function with respect to x. The function is given as \( f(x, y) = y - e^{x} \). Differentiating term by term with respect to x, we get \( f_{x} = \frac{d}{dx}(y) - \frac{d}{dx}(e^{x}) = 0 - e^{x} = -e^{x} \).

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

Next, we find the first partial derivative of the function with respect to y. Using the function \( f(x, y) = y - e^{x} \), differentiate it with respect to y: \( f_{y} = \frac{d}{dy}(y) - \frac{d}{dy}(e^{x}) = 1 - 0 = 1 \).

Step 3: Calculate the Second Partial Derivative \( f_{xx} \)

Now that we have \( f_{x} = -e^{x} \), we find \( f_{xx} \) by differentiating \( f_{x} \) again with respect to x: \( f_{xx} = \frac{d}{dx}(-e^{x}) = -e^{x} \).

Step 4: Calculate the Mixed Partial Derivative \( f_{xy} \)

Next, to find \( f_{xy} \), we take the partial derivative of \( f_{x} = -e^{x} \) with respect to y: \( f_{xy} = \frac{d}{dy}(-e^{x}) = 0 \) since \( -e^{x} \) does not depend on y.

Step 5: Calculate the Mixed Partial Derivative \( f_{yx} \)

For \( f_{yx} \), we take the partial derivative of \( f_{y} = 1 \) with respect to x: \( f_{yx} = \frac{d}{dx}(1) = 0 \), as the derivative of a constant with respect to any variable is zero. By Clairaut's Theorem, \( f_{xy} = f_{yx} \), confirming our previous result.

Step 6: Calculate the Second Partial Derivative \( f_{yy} \)

Finally, we calculate \( f_{yy} \) by differentiating \( f_{y} = 1 \) with respect to y again: \( f_{yy} = \frac{d}{dy}(1) = 0 \).

Key Concepts

Second Order Partial Derivatives

In calculus, second order partial derivatives involve taking the derivative of a function twice, either with respect to the same variable or different ones. Think of this like peeling an onion layer by layer to understand how the function changes in different dimensions, providing insights into the function's curvature.
For example, in the function given by \( f(x, y) = y - e^x \), the task is to find derivatives such as \( f_{xx}, f_{yy}, f_{xy}, \) and \( f_{yx} \).

Here's how we understand them:

• \( f_{xx} \) is the second partial derivative of \( f \), first with respect to \( x \) and then again with respect to \( x \), showing how the function changes specifically along the \( x \) direction when looked at twice.
• \( f_{yy} \) is the second derivative with respect to \( y \), highlighting changes strictly along the \( y \) direction, a simpler case in our function as it becomes zero reflecting no dependency.

These derivatives provide a snapshot of different dimensions, useful in fields like physics and engineering to model phenomena accurately.

Mixed Partial Derivatives

Mixed partial derivatives occur when you differentiate first with respect to one variable and then with another, say \( x \) and \( y \). They are like checking interactions between two different directions in a function.
In the exercise, for \( f(x, y) = y - e^x \), the mixed derivatives are:

• \( f_{xy} \): First differentiate \( f \) with respect to \( x \) (resulting in \( -e^x \)), and then differentiate this with respect to \( y \). Since \( -e^x \) is independent of \( y \), \( f_{xy} = 0 \).
• \( f_{yx} \): This involves first differentiating \( f \) with respect to \( y \) (giving \( 1 \)), and then differentiating this with respect to \( x \). As \( 1 \) is a constant, its derivative is zero, so \( f_{yx} = 0 \).

Mixed partials often show symmetry, an important concept leading us to Clairaut's Theorem.

Clairaut's Theorem

Clairaut's Theorem is a fundamental tool that asserts the consistency of mixed partial derivatives under certain conditions. It says that for functions where partial derivatives are continuous, the order of differentiation does not change the result.
Applying it to our example, with \( f(x, y) = y - e^x \), since \( f_{xy} = 0 \) and \( f_{yx} = 0 \), they indeed match, showing that the theorem holds.

This theorem is vital because it enables simplification in complex calculations, ensuring that mixed partials yield the same result regardless of the order. It’s used broadly in mathematical modeling and problem-solving, providing reassurance that the function's behavior is predictable under these conditions.