Problem 17 - Calculus and its applications

Question

Differentiate. $$ g(x)=\log _{32}(9 x-2) $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = \frac{9}{(9x-2)\ln(32)} \).

Step 1: Apply the Change of Base Formula

The function given is in the form of a logarithm with an arbitrary base, which is not natural or base 10. We can use the change of base formula to rewrite the logarithm with a different base: \[ g(x) = rac{ ext{ln}(9x-2)}{ ext{ln}(32)} \]This will make it easier to differentiate the function using properties of the natural logarithm.

Step 2: Differentiate with Respect to x

Using the rewritten form of the function, \( g(x) = \frac{\ln(9x-2)}{\ln(32)} \), apply the chain rule to differentiate: \[ g'(x) = \frac{1}{\ln(32)} \cdot \frac{d}{dx}[\ln(9x-2)] \]\( \ln(32) \) is a constant, so it remains outside the derivative. Apply the derivative of \( \ln(u) \), where \( u=9x-2 \): \( \frac{d}{dx}[\ln(9x-2)] = \frac{1}{9x-2} \cdot \frac{d}{dx}(9x-2) \).

Step 3: Differentiate the Inside Function

Inside the parentheses, \( u = 9x - 2 \). Differentiate \( u \) with respect to \( x \):\[ \frac{d}{dx}(9x - 2) = 9 \]Now incorporate this into the differentiation of the logarithm:

Step 4: Combine Derivatives to Find g'(x)

Combine all the derivatives calculated:\[ g'(x) = \frac{1}{\ln(32)} \cdot \frac{1}{9x-2} \cdot 9 \]Simplify the expression to find the derivative:\[ g'(x) = \frac{9}{(9x-2)\ln(32)} \]

Key Concepts

Understanding Change of Base Formula

When dealing with logarithms, sometimes the base isn't practical for certain calculations. This is where the change of base formula comes in handy. The formula is expressed as: \[ \log_{b}(a) = \frac{\ln(a)}{\ln(b)} \] This formula allows you to convert a logarithm to a base that is more convenient, commonly to base 10 or the natural logarithm base \( e \). This is useful because calculators and mathematical software typically handle logarithms in these bases more easily.

For example, to convert \( \log_{32}(9x - 2) \), we use \( \ln \) (natural log) resulting in:

• \( \log_{32}(9x-2) = \frac{\ln(9x-2)}{\ln(32)} \)

Conveniently, you can now work with derivatives of logarithmic functions since the natural log has well-known differentiation rules.

Exploring Natural Logarithms

Natural logarithms are logarithms with a base of \( e \), where \( e \approx 2.71828 \). They are denoted as \( \ln(x) \). Natural logs simplify calculus operations, especially differentiation.

To differentiate \( \ln(u) \), remember the rule:

• If \( u \) is a function of \( x \), then \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \)

This makes it easier to manipulate functions and take derivatives. Consider the inside function \( 9x-2 \) in \( \ln(9x-2) \). The derivative \( \frac{d}{dx}(9x-2) = 9 \), simplifies calculating \( \frac{d}{dx} \ln(9x-2) \) into:

• \( \frac{1}{9x-2} \cdot 9 \)

Thus, differentiation with natural logs follows a straightforward routine.

Applying the Chain Rule

Differentiating composite functions often requires the use of the chain rule. The chain rule provides a technique for finding the derivative of a composite function. It is expressed as:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

In the context of \( g(x) = \frac{\ln(9x-2)}{\ln(32)} \), differentiate \( \ln(9x-2) \) first using the rule for natural logarithms.

The chain rule helps combine these steps by considering outer and inner functions. First, factor out constants like \( \ln(32) \):

• \( \frac{1}{\ln(32)} \cdot \) \( \frac{1}{9x-2} \cdot 9 \)

This product represents the derivative of the entire expression. The chain rule facilitates breaking down complex derivatives into manageable parts, ensuring accuracy in differentiation of multi-layered functions.