Problem 41 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(37-58\) , find the derivative of the function. (Hint: In some exercises, you may find it helpful to apply logarithmic properties before differentiating.) $$ f(x)=x 9^{x} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = x9^{x}\) is \(f'(x) = 9^{x}(1+ x \ln(9))\).

Step 1: Identify the sub-functions

We first identify the two sub-functions that make up the function \(f(x) = x9^{x}\). These are \(u(x) = x\) and \(v(x) = 9^{x}\). The derivative of these functions will be needed for the application of the product rule.

Step 2: Apply the product rule

We can now apply the product rule, which states that the derivative of \(f(x) = u(x) \cdot v(x)\) is given by \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\). To do this, we need to find \(u'(x)\) and \(v'(x)\) (the derivatives of our sub-functions). The derivative of \(u(x) = x\) is \(u'(x) = 1\), and the derivative of \(v(x) = 9^{x}\) is found using the chain rule and the derivative of an exponential function to give \(v'(x) = 9^{x} \ln(9)\). This gives us \(f'(x) = 1 \cdot 9^{x} + x \cdot 9^{x} \ln(9)\).

Step 3: Simplify the expression

We now simplify the expression for \(f'(x)\). Notice that both terms are multiples of \(9^{x}\), so we can factor this out to give \(f'(x) = 9^{x}(1+ x \ln(9))\). This is the derivative of the function \(f(x) = x9^{x}\).

Key Concepts

Product Rule: Unraveling the Basics

When you're tasked with finding the derivative of a function that is the product of two simpler functions, this is where the product rule shines. It provides a straightforward method to differentiate functions like \( f(x) = u(x) \cdot v(x) \), where both \( u(x) \) and \( v(x) \) are themselves functions of \( x \). The product rule states that the derivative of such a function is given by:

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

This formula essentially tells us that to find the derivative, we take the derivative of the first function \( u(x) \), multiply it by the second function \( v(x) \), and then add to it the first function \( u(x) \) multiplied by the derivative of the second function \( v(x) \).
This rule is particularly useful in exercises involving functions like \( f(x) = x9^{x} \) because applying the product rule simplifies the process. It's worth noting that recognizing when to use this rule is key, especially when a function is presented as a multiplication of two or more expressions.

Chain Rule: Interacting Layers of Functions

The chain rule is fundamental when dealing with composite functions. If you have a function inside another function, this is where the chain rule helps you dissect the problem. A function like \( v(x) = 9^{x} \) involves an exponential component, where the chain rule becomes particularly handy.
In simple terms, the chain rule states that the derivative of a composite function \( h(x) = g(f(x)) \) is the derivative of the outer function \( g \) evaluated at \( f(x) \), multiplied by the derivative of the inner function \( f(x) \). Represented symbolically, this is:

• \( h'(x) = g'(f(x)) \cdot f'(x) \)

It sounds complex, but essentially you are peeling away the layers by dealing with each part of the function step-by-step.
In the context of the original problem, the derivative of \( 9^{x} \) involves recognizing \( 9^{x} \) as a composition because it can be broken down into simpler parts using the natural logarithm convention. The application of the chain rule here allows for the expression \( v'(x) = 9^{x} \ln(9) \).

Exponential Functions: Of Powers and Growth

Exponential functions are a critical aspect of calculus. They describe processes that grow or decay at a constant rate. Such functions are represented in the form \( a^{x} \), where \( a \) is a constant base, and \( x \) is the exponent.
Finding derivatives involving exponential functions typically entails using both the natural logarithm function and the chain rule. For instance, in the case of \( 9^{x} \), the derivative involves both recognizing the base and applying the natural logarithm. This is generally expressed as:

• The derivative of \( a^{x} \) is \( a^{x} \ln(a) \)

This formula shows that the derivative of an exponential function is proportional to the original function itself, with additional scaling by the natural logarithm of the base.
Understanding exponential functions in calculus helps in a variety of applications such as calculating interest, population dynamics, and predicting economic growth, emphasizing the broad significance and utility of mastering this concept.