Problem 32 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the first partial derivatives of the function at the given point. \(f(x, y)=e^{x} \ln y ;(0, e)\)

Step-by-Step Solution

Answer

In conclusion, the first partial derivatives at the point \((0, e)\) are: \[\frac{\partial f}{\partial x}(0, e) = 1\] \[\frac{\partial f}{\partial y}(0, e) = \frac{1}{e}\]

Step 1: Find the partial derivative of the function with respect to x

First, we need to find the partial derivative of the function with respect to x. To do that, treat \(y\) as a constant and differentiate the function with respect to \(x\): \[\frac{\partial f}{\partial x} = \frac{\partial (e^x \ln y)}{\partial x}\] Using the product rule for differentiation, and the fact that the derivative of \(e^x\) with respect to x is \(e^x\), we have: \[\frac{\partial f}{\partial x} = e^x \ln y \frac{\partial e^x}{\partial x} + e^x \frac{\partial \ln y}{\partial x}\] The first term becomes \(e^x \ln y\), and since \(\ln y\) is independent of x, the second term becomes 0. Therefore, we get: \[\frac{\partial f}{\partial x} = e^x \ln y\]

Step 2: Evaluate the partial derivative with respect to x at the given point

Now, we will evaluate the partial derivative with respect to x at the given point \((0, e)\): \[\frac{\partial f}{\partial x}(0, e) = e^0 \ln e\] Since \(e^0 = 1\) and \(\ln e = 1\), we have: \[\frac{\partial f}{\partial x}(0, e) = 1 \cdot 1 = 1\]

Step 3: Find the partial derivative of the function with respect to y

Next, we need to find the partial derivative of the function with respect to \(y\). To do that, treat \(x\) as a constant and differentiate the function with respect to \(y\): \[\frac{\partial f}{\partial y} = \frac{\partial (e^x \ln y)}{\partial y}\] Using the product rule for differentiation, and the fact that the derivative of \(\ln y\) with respect to y is \(\frac{1}{y}\), we have: \[\frac{\partial f}{\partial y} = e^x \ln y \frac{\partial e^x}{\partial y} + e^x \frac{\partial \ln y}{\partial y}\] The first term becomes 0, as e^x is independent of y, and the second term becomes \(\frac{e^x}{y}\). Therefore, we get: \[\frac{\partial f}{\partial y} = \frac{e^x}{y}\]

Step 4: Evaluate the partial derivative with respect to y at the given point

Finally, we will evaluate the partial derivative with respect to y at the given point \((0, e)\): \[\frac{\partial f}{\partial y}(0, e) = \frac{e^0}{e}\] Since \(e^0 = 1\), we have: \[\frac{\partial f}{\partial y}(0, e) = \frac{1}{e}\] In conclusion, the first partial derivatives at the point \((0, e)\) are: \[\frac{\partial f}{\partial x}(0, e) = 1\] \[\frac{\partial f}{\partial y}(0, e) = \frac{1}{e}\]

Key Concepts

Partial Derivatives

Calculus often deals with how things change. Partial derivatives help us to understand change in multi-variable functions. When dealing with a function like \(f(x, y)\), partial derivatives allow us to see how the function changes when we alter one variable and keep the other constant.
For example:

• To find the partial derivative of a function with respect to \(x\), treat \(y\) as a constant.
• To find the partial derivative with respect to \(y\), treat \(x\) as a constant.

This concept is crucial in fields like physics, economics, and engineering, where functions often depend on multiple variables.

Product Rule

The product rule is a fundamental technique in differentiation. It helps us find the derivative of a product of two functions. If you have two functions, say \(u(x)\) and \(v(x)\), their product's derivative is given by:\[(uv)' = u'v + uv'\] This means you take the derivative of the first function and multiply it by the second function, then add the first function multiplied by the derivative of the second. Simple!

The product rule is essential when working with functions that don’t separate easily. It ensures you differentiate correctly, saving you from potentially messy math mistakes. Don't forget to apply it when working with partial derivatives as seen in the exercise.

Differentiation

Differentiation is the process of finding a derivative. It tells us how a function is changing at any given point. You may think of it as zooming in on a curve to see how steep or flat it is at a precise location.

• A big portion of differentiation involves finding derivatives using rules like the product rule, chain rule, or quotient rule.
• When differentiating multi-variable functions, partial differentiation comes into play.

Differentiation is foundational to calculus and is used extensively in various fields like computer science for algorithms, or biology for growth rates. It’s everywhere you look, providing insights and solutions when analyzing changes.

Mathematical Functions

Mathematical functions are equations that describe relationships between variables. In calculus, these functions often require evaluation through calculus operations like integration and differentiation.

In multi-variable calculus, we focus on how these functions behave when multiple inputs are involved, such as \(f(x, y)\). These denote:

• \(x\) and \(y\) are independent variables.
• The function \(f(x, y)\) is the dependent variable.

Functions are like blueprints in math, allowing us to plot graphs, model real-world situations, or perform critical calculations. Mastery of functions and their behaviors is key to success in advanced math and related disciplines.