Problem 31 - Calculus and its applications

Question

Differentiate. $$ g(x)=x^{5} e^{2 x} $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = e^{2x}(5x^4 + 2x^5) \).

Step 1: Identify the Functions

In the function \( g(x) = x^5 e^{2x} \), identify the two functions involved: \( f(x) = x^5 \) and \( h(x) = e^{2x} \). This is a product of two functions.

Step 2: Apply the Product Rule

To differentiate the product of two functions, use the product rule: \( (uv)' = u'v + uv' \). Here \( u = x^5 \) and \( v = e^{2x} \).

Step 3: Differentiate \( u = x^5 \)

Differentiate \( u = x^5 \) with respect to \( x \): \( u' = 5x^4 \).

Step 4: Differentiate \( v = e^{2x} \)

Differentiate \( v = e^{2x} \) with respect to \( x \). Use the chain rule: \( v' = 2 e^{2x} \), because \( e^{2x} \) is an exponential function with an inner function \( 2x \), whose derivative is 2.

Step 5: Apply the Product Rule Formula

Substitute \( u' \), \( v \), \( u \), and \( v' \) into the product rule: \( g'(x) = (5x^4)(e^{2x}) + (x^5)(2e^{2x}) \).

Step 6: Simplify the Expression

Simplify the expression by factoring out the common term \( e^{2x} \): \( g'(x) = e^{2x}(5x^4 + 2x^5) \).

Key Concepts

Understanding the Product Rule

When dealing with the differentiation of products of two functions, the product rule comes handy. It helps you navigate through the mathematical wilderness of complex products by breaking them down into manageable pieces. The product rule states that the derivative of a product of two functions, say \( u(x) \) and \( v(x) \), is given by the formula:

• \((uv)' = u'v + uv'\).

This means you first take the derivative of the first function \( u(x) \), and multiply it by the second function \( v(x) \) still in its original form. Then, do the opposite—keep \( u(x) \) in its original form and multiply it by the derivative of \( v(x) \). Finally, sum these two products together. This rule is like your multi-tool when differentiating products, ensuring each component gets its spotlight without any piece being ignored.
In the context of the exercise \( g(x) = x^5 e^{2x} \), you identify \( f(x) = x^5 \) and \( h(x) = e^{2x} \) as \( u \) and \( v \) respectively. This formula streamlines seemingly complex derivatives into approachable steps.
Remember, the key strength of the product rule lies in simplifying and accurately differentiating two intertwined functions working as one.

Exploring the Chain Rule

The chain rule is a powerful tool when you need to differentiate compositions of functions. Imagine you have a function \( h(x) \) inside another function \( f \). The chain rule tells you how to find the derivative of this composite function, which you denote as \( f(g(x)) \). The chain rule states:

• \((f \, o \, g)'(x) = f'(g(x)) \cdot g'(x)\).

The chain rule allows you to break the process into digestible steps: first differentiate the outer function \( f \) with respect to the inner function \( g \), then multiply by the derivative of the inner function \( g \). In simpler terms, it's like peeling an onion: layer by layer.
Applying this to the original exercise, look at \( h(x) = e^{2x} \). Here, the outer function is the exponential, \( e^u \), and the inner function is \( u = 2x \). Differentiating \( h(x) \) requires the chain rule because you must respect the structure of the inner function driving changes in \( h \).
Like a key unlocking different doors, the chain rule reveals how nested functions influence each other while keeping the complexity in check.

Demystifying the Exponential Function

Exponential functions might seem intimidating at first glance, but they're among the most friendly functions to differentiate. An exponential function takes the form \( f(x) = a^x \) or the natural exponential, \( e^x \), where \( e \) is a mathematical constant approximately equal to 2.71828. A standout feature of exponential functions is their derivative: the derivative of \( e^x \) is simply \( e^x \) itself.

• The function doesn't 'change' when differentiated.

This makes calculations quite straightforward. However, when you encounter a transformed version, such as \( e^{gx} \), you'll need the chain rule as mentioned earlier.
In the function \( e^{2x} \), found in the original exercise, the inner function \( 2x \) changes how the exponent grows. Thus, when differentiating, you apply both the chain rule and the intrinsic property of the exponential function itself:

• \( rac{d}{dx}[e^{2x}] = 2e^{2x} \).

Remember, exponential functions appear across various scientific fields, making an understanding of their nature and differentiation a valuable asset.
With their remarkable property of retaining form through differentiation, exponential functions become less daunting and more predictable, helping you harness their potential fully.