Problem 29 - Calculus and its applications

Question

Find the four second-order partial derivatives. $$f(x, y)=7 x y^{2}+5 x y-2 y$$

Step-by-Step Solution

Answer

\( \frac{\partial^2 f}{\partial x^2} = 0\), \( \frac{\partial^2 f}{\partial y^2} = 14x\), \( \frac{\partial^2 f}{\partial y \partial x} = 14y + 5\), \( \frac{\partial^2 f}{\partial x \partial y} = 14y + 5\).

Step 1: Compute First Partial Derivative with respect to x

Given function: \(f(x, y) = 7xy^2 + 5xy - 2y\).Find \( \frac{\partial f}{\partial x} \):- Differentiate \(7xy^2\) with respect to \(x\) to get \(7y^2\).- Differentiate \(5xy\) with respect to \(x\) to get \(5y\).- Differentiate \(-2y\) with respect to \(x\), which gives \(0\) since it doesn't depend on \(x\).Thus, \( \frac{\partial f}{\partial x} = 7y^2 + 5y \).

Step 2: Compute First Partial Derivative with respect to y

Find \( \frac{\partial f}{\partial y} \):- Differentiate \(7xy^2\) with respect to \(y\) to get \(14xy\).- Differentiate \(5xy\) with respect to \(y\) to get \(5x\).- Differentiate \(-2y\) with respect to \(y\) to get \(-2\).Thus, \( \frac{\partial f}{\partial y} = 14xy + 5x - 2 \).

Step 3: Compute Second Partial Derivative with respect to x twice

Now, find \( \frac{\partial^2 f}{\partial x^2} \):Differentiate \( \frac{\partial f}{\partial x} = 7y^2 + 5y \) with respect to \(x\):- Since there are no \(x\) terms in \(7y^2 + 5y\), the derivative is \(0\).Thus, \( \frac{\partial^2 f}{\partial x^2} = 0 \).

Step 4: Compute Mixed Partial Derivative \( \frac{\partial^2 f}{\partial y \partial x} \)

Differentiate \( \frac{\partial f}{\partial x} = 7y^2 + 5y \) with respect to \(y\):- Differentiate \(7y^2\) with respect to \(y\) to get \(14y\).- Differentiate \(5y\) with respect to \(y\) to get \(5\).Thus, \( \frac{\partial^2 f}{\partial y \partial x} = 14y + 5 \).

Step 5: Compute Second Partial Derivative with respect to y twice

Find \( \frac{\partial^2 f}{\partial y^2} \):Differentiate \( \frac{\partial f}{\partial y} = 14xy + 5x - 2 \) with respect to \(y\):- Differentiate \(14xy\) with respect to \(y\) to get \(14x\).- \(5x\) does not depend on \(y\), so its derivative is \(0\).- \(-2\) with respect to \(y\) gives \(0\).Thus, \( \frac{\partial^2 f}{\partial y^2} = 14x \).

Step 6: Compute Mixed Partial Derivative \( \frac{\partial^2 f}{\partial x \partial y} \)

Differentiate \( \frac{\partial f}{\partial y} = 14xy + 5x - 2 \) with respect to \(x\):- Differentiate \(14xy\) with respect to \(x\) to get \(14y\).- Differentiate \(5x\) with respect to \(x\) to get \(5\).- \(-2\) does not depend on \(x\), so its derivative is \(0\).Thus, \( \frac{\partial^2 f}{\partial x \partial y} = 14y + 5 \).

Key Concepts

Understanding First Partial Derivatives

When we talk about first partial derivatives, we're examining how a multivariable function changes as we adjust one variable while keeping the others constant. Consider the function provided: \( f(x, y) = 7xy^2 + 5xy - 2y \). This is a function of two variables, \( x \) and \( y \). First partial derivatives help us to understand how these variables individually influence the output of the function.

To find the first partial derivative with respect to \( x \), we treat \( y \) as a constant and differentiate only with respect to \( x \). Here's how it's done:

• For \( 7xy^2 \), differentiating with respect to \( x \) yields \( 7y^2 \)
• Similarly, for the term \( 5xy \), the derivative is \( 5y \)
• The term \(-2y\) has no \( x \) in it, so its derivative is \( 0 \)

Hence, the first partial derivative of \( f \) with respect to \( x \) is \( 7y^2 + 5y \).

Similarly, to find the first partial derivative with respect to \( y \), we differentiate by treating \( x \) as a constant:

• For \( 7xy^2 \), getting the derivative gives us \( 14xy \)
• For \( 5xy \), it becomes \( 5x \)
• And \(-2y\) simply becomes \(-2\)

Thus, with respect to \( y \), the derivative is \( 14xy + 5x - 2 \). These first derivatives are fundamental because they reveal instantaneous rates of change in separate directions.

Exploring Second-Order Partial Derivatives

Second-order partial derivatives give us deeper insights about how a function behaves by describing how the rate of change itself changes. More specifically, for the function \( f(x, y) = 7xy^2 + 5xy - 2y \), these derivatives will tell us about concavity and can help determine maxima, minima, or saddle points in multivariable calculus related problems.

First, let's look at the second partial derivative with respect to \( x \) twice, denoted by \( \frac{\partial^2 f}{\partial x^2} \). When we differentiate \( 7y^2 + 5y \) again with respect to \( x \), both terms are constants relative to \( x \), giving a result of \( 0 \).
For the second partial derivative with respect to \( y \) twice, denoted by \( \frac{\partial^2 f}{\partial y^2} \), we differentiate \( 14xy + 5x - 2 \) with respect to \( y \) giving us \( 14x \).

These derivative calculations emphasize how fixed certain aspects of the function are with respect to changes in a single variable.

• \( \frac{\partial^2 f}{\partial x^2} \) tells us that along the \( x \) direction, the function's curvature doesn't change.
• \( \frac{\partial^2 f}{\partial y^2} = 14x \) implies that the function's curvature changes linearly with \( x \) along the \( y \) direction.

Detecting where these values are zero could hint at critical points where significant variations in the function's behavior might occur.

Deciphering Mixed Partial Derivatives

Mixed partial derivatives are a bit more nuanced. They look at how a function changes when both variables are altered. In practical terms, these derivatives can tell us about the interaction between two variables in influencing a function’s behavior.

For our function \( f(x, y) = 7xy^2 + 5xy - 2y \), we consider two mixed partial derivatives:

• \( \frac{\partial^2 f}{\partial y \partial x} \)
• \( \frac{\partial^2 f}{\partial x \partial y} \)

Interestingly, for many functions, these two will be equal due to what's known as Clairaut's theorem, provided the function is well-behaved and differentiable in its domain.

First, we found \( \frac{\partial^2 f}{\partial y \partial x} = 14y + 5 \) by differentiating \( 7y^2 + 5y \) again with respect to \( y \), showing how changes in \( y \) affect changes already made in \( x \). Similarly, \( \frac{\partial^2 f}{\partial x \partial y} = 14y + 5 \) confirms this.

• Both derivatives illustrate the interaction effect between \( x \) and \( y \) at every point along the input space.
• When mixed derivatives exist and equate, this asserts a symmetry in the way these changes affect \( f \).

This symmetry is crucial, providing a consistent way to explore and predict multivariable function behavior. Identifying and understanding these mixed derivatives paves the way for more advanced investigations and applications, such as analyzing surfaces and modeling real-world systems.