Problem 31 - APEX Calculus

Question

Find \(f_{x}, f_{y}, f_{z}, f_{y z}\) and \(f_{z y}\). $$ f(x, y, z)=x^{2} e^{2 y-3 z} $$

Step-by-Step Solution

Answer

\( f_x = 2x e^{2y-3z}, f_y = 2x^2 e^{2y-3z}, f_z = -3x^2 e^{2y-3z}, f_{yz} = f_{zy} = -6x^2 e^{2y-3z} \).

Step 1: Find the Partial Derivative with Respect to x

To find \( f_x \), treat \( y \) and \( z \) as constants and differentiate \( f(x, y, z) = x^2 e^{2y - 3z} \) with respect to \( x \). The derivative is: \[ f_x = \frac{\partial}{\partial x} \left( x^2 e^{2y - 3z} \right) = 2x e^{2y - 3z}. \]

Step 2: Find the Partial Derivative with Respect to y

To find \( f_y \), treat \( x \) and \( z \) as constants and differentiate \( f(x, y, z) = x^2 e^{2y - 3z} \) with respect to \( y \). The derivative is: \[ f_y = \frac{\partial}{\partial y} \left( x^2 e^{2y - 3z} \right) = x^2 \cdot 2e^{2y - 3z} = 2x^2 e^{2y - 3z}. \]

Step 3: Find the Partial Derivative with Respect to z

To find \( f_z \), treat \( x \) and \( y \) as constants and differentiate \( f(x, y, z) = x^2 e^{2y - 3z} \) with respect to \( z \). The derivative is: \[ f_z = \frac{\partial}{\partial z} \left( x^2 e^{2y - 3z} \right) = x^2 (-3)e^{2y - 3z} = -3x^2 e^{2y - 3z}. \]

Step 4: Find the Mixed Partial Derivative with Respect to y then z

To find \( f_{yz} \), first find \( f_y = 2x^2 e^{2y - 3z} \) from Step 2, and then differentiate with respect to \( z \). The derivative is: \[ f_{yz} = \frac{\partial}{\partial z} (2x^2 e^{2y - 3z}) = 2x^2 (-3) e^{2y - 3z} = -6x^2 e^{2y - 3z}. \]

Step 5: Find the Mixed Partial Derivative with Respect to z then y

To find \( f_{zy} \), first find \( f_z = -3x^2 e^{2y - 3z} \) from Step 3, and then differentiate with respect to \( y \). The derivative is: \[ f_{zy} = \frac{\partial}{\partial y} (-3x^2 e^{2y - 3z}) = -3x^2 (2) e^{2y - 3z} = -6x^2 e^{2y - 3z}. \]

Key Concepts

Multivariable Calculus

Multivariable calculus is an extension of calculus that deals with functions of multiple variables. It's essential for understanding how changes in one variable can affect others.
Here's a breakdown of key elements:

• **Functions of Several Variables**: These are functions that have more than one input. For example, a temperature reading might depend on latitude, longitude, and altitude - three variables.
• **Partial Derivatives**: These are just regular derivatives, but taken with regard to one variable at a time while keeping others constant.
• **Gradients and Directional Derivatives**: These tools show how functions change with respect to all variables, not just one.

In the problem we've seen, multivariable calculus was used to find how each variable impacts the function by calculating its partial derivatives.

Partial Differentiation

Partial differentiation is a technique of finding the derivative of a function concerning one particular variable, while treating other variables as constants. Let's look at how this works:

• **Focus on One Variable at a Time**: In the function \( f(x, y, z) = x^2 e^{2y - 3z} \), we differentiate once with respect to each variable: \( x \), \( y \), and \( z \).
• **Keeping Other Variables Constant**: For \( f_x \), treat \( y \) and \( z \) as constants. Similarly, for \( f_y \), \( x \) and \( z \) are constants.
• **Using Rules of Differentiation**: Differentiate using familiar rules, such as the power rule and chain rule, adapted for functions of several variables.

The exercise involved finding \( f_x, f_y, \) and \( f_z \), illustrating how output behavior changes as each variable shifts, still holding others fixed.

Mixed Partial Derivatives

Mixed partial derivatives provide insight into how changes in two variables, together, affect a function. These derivatives are taken one after the other concerning different variables:

• **Stepwise Approach**: First, perform partial differentiation with respect to one variable, then repeat for another. Sequential differentiation like this is typical in calculating mixed partial derivatives.
• **Order Doesn't Always Matter**: For many functions, the second-order mixed derivatives \( f_{yz} \) and \( f_{zy} \) are equal due to Clairaut's theorem, assuming continuous second derivatives. Our exercise showed \( f_{yz} = f_{zy} = -6x^2 e^{2y - 3z} \).
• **Practical Application**: Understanding these derivatives helps in analyzing how two changes affect the overall system behavior, crucial in fields like economics, physics, and engineering.

This exercise emphasized the process of applying mixed partial derivatives to understand the relationship between changes in \( y \) and \( z \). This is particularly useful in scenarios where multiple factors occur simultaneously.