Problem 24 - Calculus for Biology and Medicine

Question

Find the indicated partial derivatives. \(f(u, v)=e^{\mu+3 v^{2}} ; f_{u}(2,1)\)

Step-by-Step Solution

Answer

The partial derivative \(f_u(2,1)\) is approximately 148.413.

Step 1: Differentiate with respect to u

First, compute the partial derivative of the function with respect to the variable \(u\). Doing this requires treating \(v\) as a constant. The original function is \(f(u, v) = e^{u + 3v^2}\). Differentiating with respect to \(u\), the derivative \(f_u\) is given by the chain rule: since the derivative of \(e^x\) with respect to \(x\) is \(e^x\), we have: \[ f_u(u, v) = \frac{\partial}{\partial u} [e^{u + 3v^2}] = e^{u + 3v^2} \]

Step 2: Substitute the values u=2 and v=1

Now, substitute the given point \((u, v) = (2, 1)\) into the partial derivative expression found in Step 1. Replace \(u\) with 2 and \(v\) with 1:\[ f_u(2, 1) = e^{2 + 3(1)^2} = e^{2+3} = e^5 \]

Step 3: Compute the final result

Compute the numerical value of \(e^5\). Since \(e\) is the base of the natural logarithm, it's approximately equal to 2.71828. Thus, the numerical value of \(e^5\) is roughly:\[ e^5 \approx 2.71828^5 \approx 148.413 \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that helps us differentiate composite functions. Imagine you have a function nested within another function, like how function - takes an input, does something to it, and then outputs a result that gets further processed by another function. The chain rule tells us how to find the derivative of this kind of composite function.
To apply the chain rule, you:

• First, take the derivative of the outer function while keeping the inner function unchanged.
• Then, multiply this result by the derivative of the inner function.

In the exercise, the chain rule was applied to the function \( f(u, v) = e^{u + 3v^2} \).
With respect to \( u \), the derivative of \( e^x \) where \( x = u + 3v^2 \), remains \( e^x \), showing how we handled the outer function. Since the derivative of \( u \) is simply 1, the result simplifies nicely. This is why applying the chain rule became straightforward here, even though it involves exponential functions as the outer layer.

Exponential Functions

An exponential function is a type of mathematical function where a constant base is raised to a variable exponent. These functions have the form \( f(x) = a^x \), where \( a \) is a positive constant.
In calculus, the most common exponential function encountered is \( e^x \), where \( e \approx 2.71828 \), known as Euler's number. It's unique because the derivative of \( e^x \) is \( e^x \) itself. This property simplifies the differentiation process, particularly when dealing with exponential growth or decay models.
In our exercise, the function \( f(u, v) = e^{u + 3v^2} \) highlights this property. When we differentiate with respect to \( u \), especially using the chain rule, this property ensures we got \( e^{u + 3v^2} \) as the derivative. Recognizing and working with exponential functions becomes handy in solving more complex calculus problems efficiently.

Calculus Problems

Calculus problems often require a blend of different calculus concepts to find solutions. The goal is often to understand the rate at which things change. A consistent strategy in solving these problems involves knowing which rule or concept to apply.
In calculus, differentiation is critical for finding rates of change. Partial derivatives, specifically, allow us to focus on how one variable affects the function when others are held constant. This was exactly the task in the given exercise: you found the partial derivative of \( f(u, v) = e^{u + 3v^2} \) with respect to \( u \), treating \( v \) as constant.
When tackling calculus problems, it helps to:

• Clearly identify what you need to find – here, a specific partial derivative.
• Apply the appropriate rules; for partial derivatives, this often involves holding other variables constant.
• Substitute any given values to find numerical results.

Such steps make even more complex problems manageable, offering a clear path from understanding the question to finding a solution.