Problem 10 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=2 x e^{-3 x} $$

Step-by-Step Solution

Answer

\((2 - 6x)e^{-3x}\)

Step 1: Identify the Type of Function

The given function is a product of two functions of \(x\): \(2x\) and \(e^{-3x}\). To differentiate this, we'll use the product rule.

Step 2: Write Down the Product Rule Formula

The product rule states that if you have two functions \(u(x)\) and \(v(x)\), their derivative is given by: \((uv)' = u'v + uv'\). Here, \(u(x) = 2x\) and \(v(x) = e^{-3x}\).

Step 3: Differentiate Each Function Individually

Find the derivatives of \(u(x)\) and \(v(x)\):- \(u(x) = 2x \Rightarrow u'(x) = 2\).- \(v(x) = e^{-3x} \Rightarrow v'(x) = -3e^{-3x}\). This uses the chain rule for \(v(x)\).

Step 4: Apply the Product Rule

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the product rule:\[ \frac{d}{dx}(2x \cdot e^{-3x}) = 2 \cdot e^{-3x} + 2x \cdot (-3e^{-3x}) \].

Step 5: Simplify the Expression

Simplify the expression:\[ 2e^{-3x} - 6xe^{-3x} \].Factor out the common term \(e^{-3x}\):\[ (2 - 6x)e^{-3x} \].

Key Concepts

Product Rule

The Product Rule is a fundamental concept in calculus for differentiating products of two functions. When you have a function composed as a product of two simpler functions, you can't simply differentiate each part separately and multiply them together. Instead, you need to apply the Product Rule.

The Product Rule formula is given by:

• If you have functions \(u(x)\) and \(v(x)\), the derivative of their product \( uv \) is \((uv)' = u'v + uv'\).

This means:

• First, differentiate the first function \(u(x)\) to get \(u'(x)\), while leaving \(v(x)\) alone.
• Then, do the opposite: leave \(u(x)\) as is, and differentiate \(v(x)\) to get \(v'(x)\).
• Multiply these derivatives by the opposite function and then add the results together.

Using the Product Rule allows for accurate differentiation of complex products that cannot simply be split into separate operations.

Chain Rule

When it comes to differentiating composite function systems, the Chain Rule is your go-to tool. The given function \(v(x) = e^{-3x}\) is a composition of the exponential function and a linear function.

The Chain Rule allows you to differentiate such functions systematically by addressing the outer and inner functions separately:

• Identify the outer function \(g(u)\) and the inner function \(u(x)\).
• Differentiate the outer function with respect to the inner function, meaning find \(g'(u)\).
• Differentiate the inner function as if it were a stand-alone function to find \(u'(x)\).
• The derivative of the composite function \(f(x) = g(u(x))\) is then found using \(f'(x) = g'(u) \cdot u'(x)\).

In this exercise, applying the Chain Rule simplifies finding \(v'(x) = -3e^{-3x}\), allowing us to handle complex combinations like exponentials with inner adjustments.

Exponential Functions

Understanding exponential functions, especially when they involve the natural base \(e\), is crucial in calculus. An exponential function in its simplest form can be written as \(e^{f(x)}\), where \(f(x)\) is some expression in terms of \(x\).

In our exercise, we deal with \(e^{-3x}\):

• The base \(e\) is a mathematical constant approximately equal to 2.71828, and it is the fundamental base of natural logarithms.
• The exponent, which can be any expression, determines how the exponential function rises or falls.
• When differentiating \(e^{u}\) using the Chain Rule, you multiply the original function by the derivative of the exponent. So \(d(e^{u})/dx = e^{u} \cdot u'(x)\).

Exponential functions are powerful due to their unique properties, like rapid growth, and recognizing how to differentiate them using these rules is key in calculus.

Step-by-Step Solution

Breaking down a problem step-by-step makes any task manageable. Here's a brief overview of how the step-by-step differentiation was done in this exercise which begins by identifying the function type.

First, we recognize that \(f(x) = 2x e^{-3x}\) is a product of two simpler functions. So, the Product Rule comes into play.

• We define \(u(x) = 2x\) and identify its derivative as \(u'(x) = 2\).
• We then set \(v(x) = e^{-3x}\) and find its derivative \(v'(x) = -3e^{-3x}\) using the Chain Rule.
• Substituting into the Product Rule gives: \(\frac{d}{dx}(2x \cdot e^{-3x}) = 2 \cdot e^{-3x} + 2x \cdot (-3e^{-3x})\).
• Simplifying, the derivative becomes \((2 - 6x)e^{-3x}\).

This methodical, detailed approach ensures the solution is clear and understandable, reinforcing the principles of calculus.