Problem 16 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the first partial derivatives of the function. \(f(x, y)=\left(e^{x}+e^{y}\right)^{5}\)

Step-by-Step Solution

Answer

The first partial derivatives of the function \(f(x, y) = (e^x + e^y)^5\) are: \( \frac{\partial f(x, y)}{\partial x} = 5\left(e^x + e^y\right)^4 \cdot e^x \) \( \frac{\partial f(x, y)}{\partial y} = 5\left(e^x + e^y\right)^4 \cdot e^y \)

Step 1: Differentiate with respect to x

First, we need to differentiate the function \(f(x, y)\) with respect to x, while treating y as a constant. We'll use the chain rule, as we have an outer function (exponentiation) and an inner function (sum of exponentials). \( \frac{\partial f(x, y)}{\partial x} = \frac{\partial}{\partial x} \left(e^x + e^y \right)^5 \) Using the chain rule, first differentiate the outer function, then multiply by the derivative of the inner function: \( \frac{\partial f(x, y)}{\partial x} = 5\left(e^x + e^y \right)^4 \cdot \frac{\partial}{\partial x}\left(e^x + e^y\right) \) Now, differentiate the inner sum, differentiating each term separately: \( \frac{\partial}{\partial x}\left(e^x + e^y\right) = e^x + 0 \) \( \frac{\partial}{\partial x}\left(e^x + e^y\right) = e^x \) And substitute back into our expression for the derivative: \( \frac{\partial f(x, y)}{\partial x} = 5\left(e^x + e^y\right)^4 \cdot e^x \)

Step 2: Differentiate with respect to y

Now we need to differentiate the function \(f(x, y)\) with respect to y, while treating x as a constant. We'll follow the same process as before. \( \frac{\partial f(x, y)}{\partial y} = \frac{\partial}{\partial y} \left(e^x + e^y\right)^5 \) Again, using the chain rule: \( \frac{\partial f(x, y)}{\partial y} = 5\left(e^x + e^y\right)^4 \cdot \frac{\partial}{\partial y}\left(e^x + e^y\right) \) Now, differentiate the inner sum, differentiating each term separately: \( \frac{\partial}{\partial y}\left(e^x + e^y\right) = 0 + e^y \) \( \frac{\partial}{\partial y}\left(e^x + e^y\right) = e^y \) And substitute back into our expression for the derivative: \( \frac{\partial f(x, y)}{\partial y} = 5\left(e^x + e^y\right)^4 \cdot e^y \)

Step 3: Presenting the results

Now, we have the first partial derivatives of the function: \( \frac{\partial f(x, y)}{\partial x} = 5\left(e^x + e^y\right)^4 \cdot e^x \) \( \frac{\partial f(x, y)}{\partial y} = 5\left(e^x + e^y\right)^4 \cdot e^y \)

Key Concepts

Understanding the Chain Rule in Differential Calculus

The chain rule is an essential concept in differential calculus when you're dealing with compositions of functions. Imagine you have a function nested within another, like our example: the function \( (e^x + e^y)^5 \). Here, the outer function is the power of 5, and the inner function is the expression \( e^x + e^y \). To find the derivative, you need to differentiate these functions step-by-step, applying the chain rule.The chain rule states that you differentiate the outer function while keeping the inner function constant, then multiply this by the derivative of the inner function. For our exercise:

• Differentiate the outer function: Derivative of \( u^5 \) (where \( u = e^x + e^y \)) is \( 5u^4 \).
• Multiply by the derivative of the inner function, \( e^x + e^y \).

Applying these steps correctly, we can approach more complex derivatives with confidence.
Remember, the chain rule acts as a bridge, helping you manage the relationship between nested functions effectively.

Differential Calculus Basics

Differential calculus involves studying how and why functions change—a core aspect is finding derivatives of functions. Derivatives measure the rate of change of a quantity, acting much like a mathematical microscope that examines small changes.When you calculate a derivative, you are essentially finding a slope or rate of change. This is extremely useful for predicting how a small change in one variable might impact another.

• For a function \( f(x) \), the derivative \( \frac{df}{dx} \) represents its rate of change with respect to \( x \).
• In multivariable settings, partial derivatives extend this concept to functions of more than one variable.

In our exercise, differential calculus is applied to obtain partial derivatives and explore how each variable separately affects the function. Knowing how to derive and interpret these rates of change is fundamental in various fields ranging from physics to economics. Understanding differential calculus is like acquiring a toolkit for analyzing dynamic systems.

Excursion into Multivariable Calculus

Multivariable calculus is a branch of calculus that deals with functions of multiple variables. Picture it as an extension from a single variable to several, unraveling a more complex and detailed canvas of analysis.In multivariable calculus, instead of just \( f(x) \), you have functions like \( f(x, y, z) \), which can represent more intricate systems. This requires partial derivatives to see how each variable individually affects the function, holding the other variables constant.

• Partial derivative \( \frac{\partial f}{\partial x} \) keeps \( y \) constant.
• Partial derivative \( \frac{\partial f}{\partial y} \) keeps \( x \) constant.

In our example, we saw how each variable impacts the function by calculating its respective partial derivatives. This kind of analysis is invaluable whenever you need to model and solve real-world problems that depend on several input factors simultaneously.
Grasping multivariable calculus enhances your ability to think in dimensions, adding depth to problem-solving scenarios.