Problem 10 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(f(x)=2 x e^{-3 x}\)

Step-by-Step Solution

Answer

The derivative of the function is \(f'(x) = e^{-3x}(2 - 6x)\).

Step 1: Identify the type of function

Observe that the function to be differentiated, \(f(x) = 2x e^{-3x}\), is the product of two functions: \(2x\) and \(e^{-3x}\). This means the product rule of differentiation will be needed.

Step 2: Recall the Product Rule

The product rule states that for two functions \(u(x)\) and \(v(x)\), the derivative of their product is \((uv)' = u'v + uv'\). In our case, \(u(x) = 2x\) and \(v(x) = e^{-3x}\).

Step 3: Differentiate the First Function

Differentiate \(u(x) = 2x\) with respect to \(x\). The derivative is \(u'(x) = 2\) because the coefficient 2 is constant and the derivative of \(x\) is 1.

Step 4: Differentiate the Second Function

Differentiate \(v(x) = e^{-3x}\) with respect to \(x\). Using the chain rule, \(v'(x) = e^{-3x} \cdot (-3) = -3e^{-3x}\).

Step 5: Apply the Product Rule

Substitute \(u(x), u'(x), v(x),\) and \(v'(x)\) into the product rule formula: \((2x \, e^{-3x})' = 2 \, e^{-3x} + 2x \, (-3e^{-3x})\).

Step 6: Simplify the Expression

Simplify the expression obtained from the product rule: \(2e^{-3x} - 6xe^{-3x}\). Factor out the common term \(e^{-3x}\). The result is \(e^{-3x}(2 - 6x)\).

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus, particularly when differentiating the product of two functions.
When you have two functions multiplied together, like in our original function, the product rule makes differentiation straightforward.
To recall, the product rule states that if you have functions \( u(x) \) and \( v(x) \), then the derivative of the product \( (uv)' \) is given by the formula:

• \( (uv)' = u'v + uv' \)

This means you first take the derivative of the first function \( u(x) \), then multiply it by the second function \( v(x) \), and add it to the product of the first function \( u(x) \) and the derivative of the second function \( v(x) \).
It's essential to correctly identify \( u(x) \) and \( v(x) \) in your function to apply this rule effectively. In our exercise, we identified \( u(x) = 2x \) and \( v(x) = e^{-3x} \). This straightforward approach helps ensure that you apply the product rule accurately.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change or the slope of the function at any given point. Differentiation can be applied using various rules depending on the structure of the function.
In simpler terms, differentiation helps us understand how a function changes as its input changes.
The strategies to differentiate include:

• Power Rule: Useful for terms like \( ax^n \) where the derivative becomes \( n \times ax^{n-1} \).
• Constant Rule: The derivative of a constant is zero.
• Product Rule and Chain Rule: As experienced in this exercise, useful for more complex expressions.

By differentiating the functions \( u(x) = 2x \) and \( v(x) = e^{-3x} \) separately, and then combining their results using the product rule, we can determine the derivative of the entire function effectively.

Chain Rule

The chain rule is another crucial tool in differentiation, especially when dealing with composite functions. A composite function is essentially a function within a function.
When applying the chain rule, we differentiate the outer function first and then multiply it by the derivative of the inner function.
In the context of our exercise, when differentiating \( v(x) = e^{-3x} \), we employ the chain rule.
The chain rule is essential when you have compositions like \( e^{g(x)} \) where \( g(x) \) itself is a function of \( x \).

• For \( e^{-3x} \), start by recognizing \( g(x) = -3x \).
• Differentiate the exponential \( e^{g(x)} \) which remains \( e^{g(x)} \).
• Multiply by the derivative of the inner function \( g(x) \), which is \( -3 \).

So the derivative, \( v'(x) \), results in \( -3e^{-3x} \).
Understanding the chain rule not only helps with this particular exercise but is also applicable to a wide range of complex functions you may encounter in calculus.