Problem 53 - Calculus and its applications

Question

Differentiate. $$ g(x)=\left(4 x^{2}+3 x\right) e^{x^{2}-7 x} $$

Step-by-Step Solution

Answer

Use the product rule and chain rule to find that \( g'(x) = e^{x^2-7x} \left[(8x+3) + (4x^2+3x)(2x-7) \right] \).

Step 1: Identify Differentiation Rule

The given function is of the form \( u(x) \, v(x) \), where \( u(x) = (4x^2 + 3x) \) and \( v(x) = e^{x^2 - 7x} \). Therefore, we will use the product rule for differentiation: \( g(x) = u(x) \, v(x) \rightarrow g'(x) = u'(x) \, v(x) + u(x) \, v'(x) \).

Step 2: Differentiate \( u(x) \)

Calculate the derivative of \( u(x) = 4x^2 + 3x \). For power functions, we use the rule \( \frac{d}{dx}[x^n] = nx^{n-1} \). Hence, \( u'(x) = 8x + 3 \).

Step 3: Differentiate \( v(x) \)

Calculate the derivative of \( v(x) = e^{x^2 - 7x} \) using the chain rule. The chain rule states that \( \frac{d}{dx}[e^{u}] = e^{u} \cdot \frac{du}{dx} \). Here, \( u = x^2 - 7x \), so \( \frac{du}{dx} = 2x - 7 \). Thus, \( v'(x) = e^{x^2 - 7x} \cdot (2x - 7) \).

Step 4: Apply the Product Rule

Now apply the product rule: \[ g'(x) = (8x + 3) \cdot e^{x^2 - 7x} + (4x^2 + 3x) \cdot (e^{x^2 - 7x} \cdot (2x - 7)) \].

Step 5: Simplify the Expression

Factor out \( e^{x^2 - 7x} \) from the expression: \[ g'(x) = e^{x^2 - 7x} \cdot \left[ (8x + 3) + (4x^2 + 3x)(2x - 7) \right] \]. Expand and combine like terms to get the final expression.

Key Concepts

Product Rule

In calculus, the product rule is a fundamental technique used to differentiate functions that are the product of two functions. When you have a function of the form \( u(x) \cdot v(x) \), the product rule helps to find the derivative. The formula is:

• \( g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

To apply the product rule effectively, you need to:

• Identify the two functions, \( u(x) \) and \( v(x) \).
• Compute the derivative of each function separately, namely \( u'(x) \) and \( v'(x) \).
• Substitute these derivatives back into the formula.

The product rule allows us to find the overall rate of change of the product of two varying quantities. This is crucial when dealing with complex functions in calculus problems. Below, we will see how it connects with other differentiation rules.
You will often encounter the product rule in combination with other rules like the chain rule, especially when dealing with more complex expressions such as products involving exponential functions.

Chain Rule

The chain rule is another essential tool in differentiation and is particularly useful when dealing with composite functions. This rule allows you to differentiate a function that is "inside" another function.

• If you have a function \( f(g(x)) \), the chain rule states that the derivative is: \( f'(g(x)) \cdot g'(x) \)

For example, in the differentiation of \( v(x) = e^{x^2 - 7x} \), the chain rule is applied. Here, the composite function is the exponential function where the exponent is itself a function, \( x^2 - 7x \). By using the chain rule, we differentiate it as follows:

• Let \( u = x^2 - 7x \).
• Then, find \( \frac{du}{dx} = 2x - 7 \).
• The derivative of the exponential function is \( e^u \cdot \frac{du}{dx} \).

This technique is pivotal in handling nested functions and makes solving complex differentiation problems more manageable. Understanding the chain rule is crucial because it often simplifies tasks that involve differentiating functions like natural exponentials.

Exponential Functions

Exponential functions, usually characterized by the form \( e^x \), have unique properties that make them stand out in calculus.

• The derivative of an exponential function \( e^x \) is \( e^x \), showcasing unchanged growth at any point on its curve.

These functions come with predictable behavior which makes them useful across various scientific and engineering fields.
When working with exponential functions like \( e^{x^2 - 7x} \), the differentiation process often involves combining other rules. This specific function is differentiated using the chain rule, resulting in a derivative of \( e^{x^2 - 7x} \cdot (2x - 7) \). Here, you notice how the rate of change depends both on the exponential's base \( e \) and the interior derivative \( 2x - 7 \).
Exponential functions grow at rates proportional to their current value. This makes them suitable for modeling processes with consistent percentage growth rates such as population growth, financial investments, or radioactive decay. Thus, mastering exponential differentiation helps in various practical applications.