Problem 8 - Applied Calculus

Question

Find the partial derivatives in Problems. The variables are restricted to a domain on which the function is defined. $$f_{x} \text { and } f_{y} \text { if } f(x, y)=5 x^{2} y^{3}+8 x y^{2}-3 x^{2}$$

Step-by-Step Solution

Answer

\( f_x = 10xy^3 + 8y^2 - 6x \) and \( f_y = 15x^2y^2 + 16xy \).

Step 1: Identify the Function and Variables

We are given the function \( f(x, y) = 5x^2y^3 + 8xy^2 - 3x^2 \). We need to find the partial derivatives \( f_x \) and \( f_y \), where \( f_x \) represents the partial derivative with respect to \( x \), and \( f_y \) represents the partial derivative with respect to \( y \).

Step 2: Calculate the Partial Derivative with Respect to x

To find \( f_x \), differentiate the function with respect to \( x \), treating \( y \) as a constant. This gives:\[ f_x = \frac{\partial}{\partial x}(5x^2y^3 + 8xy^2 - 3x^2) \]By applying differentiation rules, we get:\[ f_x = (10x y^3) + (8y^2) - (6x) \]

Step 3: Calculate the Partial Derivative with Respect to y

To find \( f_y \), differentiate the function with respect to \( y \), treating \( x \) as a constant. This gives:\[ f_y = \frac{\partial}{\partial y}(5x^2y^3 + 8xy^2 - 3x^2) \]By applying differentiation rules, we get:\[ f_y = (15x^2 y^2) + (16x y) \]

Step 4: Write the Final Solutions for Partial Derivatives

The partial derivative of \( f \) with respect to \( x \) is \( f_x = 10xy^3 + 8y^2 - 6x \).The partial derivative of \( f \) with respect to \( y \) is \( f_y = 15x^2y^2 + 16xy \).

Key Concepts

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of several variables. This kind of calculus goes beyond single-variable calculus, which often deals with functions of only one variable. In multivariable calculus, you're dealing with functions that take two or more variables as inputs.
Imagine trying to describe a landscape. You wouldn't just need latitude or longitude, but both, to get a full picture. That's what multivariable functions do by using multiple inputs to determine an output.
These functions help us model complex scenarios in physics, economics, and beyond, where more than one factor influences the outcome. It allows us to explore and solve problems that more accurately reflect the world around us.

• Functions can be written in the form of several variables, like in the given function \( f(x, y) \)
• You can visualize them as surfaces, enabling a deeper understanding of how changes to one variable affect another
• In real-life applications, such scenarios include temperature distribution, population dynamics, and financial modeling

Differentiation

Differentiation is a fundamental tool in calculus, allowing us to determine the rate at which a quantity changes. For single-variable functions, this involves finding the derivative with respect to one variable. However, in multivariable calculus, we extend this concept to find partial derivatives.
When we talk about partial derivatives, we're referring to the derivative of a function with more than one variable with respect to one of those variables, while keeping others constant. This is particularly useful when analyzing systems where each variable plays a different role.

• The process involves treating all other variables as constants and differentiating with respect to the chosen variable.
• For example, in the function \( f(x, y) = 5x^2y^3 + 8xy^2 - 3x^2 \), partial derivatives help distinguish how the function changes with respect to just \( x \) or just \( y \).
• Partial derivatives provide essential information in optimization and modeling by showing how each independent variable contributes to overall change.

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