Problem 18 - Calculus and its applications

Question

Differentiate. $$ g(x)=\log _{6}(5 x+1) $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = \frac{5}{(5x+1)\ln(6)} \).

Step 1: Convert Logarithm to Natural Logarithm

To differentiate a logarithm with a base other than 10 or e, we first convert it to a natural logarithm using the change of base formula. The function can be rewritten as: \[ g(x) = \frac{\ln(5x+1)}{\ln(6)} \] Since \( \ln(6) \) is a constant, this allows us to focus on differentiating \( \ln(5x+1) \).

Step 2: Differentiate using the Chain Rule

Differentiate \( g(x) = \frac{1}{ \ln(6)} \ln(5x+1) \). Use the chain rule to differentiate \( \ln(5x+1) \). The chain rule states \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \). Here \( u = 5x + 1 \), so \( \frac{du}{dx} = 5 \).

Step 3: Apply the Chain Rule

Now apply the chain rule: \[ \frac{d}{dx} \ln(5x+1) = \frac{1}{5x+1} \cdot 5 \] Therefore, \( g'(x) = \frac{1}{\ln(6)} \cdot \frac{5}{5x+1} \).

Step 4: Simplify the Derivative

Simplify the expression for the derivative: \[ g'(x) = \frac{5}{(5x+1) \ln(6)} \] This is the simplified expression of the derivative for the function \( g(x) \).

Key Concepts

Change of Base Formula

When faced with logarithms that have a base other than 10 or the natural base, \(e\), we can utilize the change of base formula to make differentiation easier. This formula is helpful because it allows us to convert any logarithmic expression into one that involves natural logarithms, which are easier to work with in calculus.

The change of base formula states:

• \( \log_b(a) = \frac{\ln(a)}{\ln(b)} \)

This means you can express a logarithm in any base, \(b\), as the quotient of two natural logarithms. In the case of our original exercise, \(g(x)=\log _{6}(5 x+1)\), we use the change of base formula to rewrite it as:

• \( g(x) = \frac{\ln(5x+1)}{\ln(6)} \)

By using the formula, we simplify our differentiation task. The division by \(\ln(6)\) makes it a constant factor, allowing us to focus on the numerator, \(\ln(5x+1)\), which is now ready for differentiation.

Chain Rule

The chain rule is a powerful differentiation tool that helps us find derivatives of composite functions. A composite function is essentially a function inside another function. The chain rule states:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

In layman's terms, you first differentiate the outer function, leaving the inner function unchanged, and then multiply by the derivative of the inner function.

In the context of our exercise, we aim to differentiate:\( g(x) = \frac{1}{\ln(6)} \ln(5x+1) \). It's evident that \(\ln(5x+1)\) is a composite function, with the inner function being \(u = 5x+1\).

By applying the chain rule, we find:

• Differentiate \(\ln(u)\) with respect to \(u\), which gives \(\frac{1}{u}\)
• Then multiply by the derivative of the inner function \(u = 5x+1\), which is \( \frac{du}{dx} = 5 \)

The result is \(\frac{1}{5x+1} \cdot 5\), leading us towards the full derivative expression.

Differentiation Techniques

Differentiating logarithmic functions involves a few distinctive techniques. First off, when differentiating logarithms of any base, such as \( g(x)=\log _{6}(5 x+1) \), converting to a natural logarithm simplifies the process. This technique reduces the problem to a standard form involving \(\ln\).

Once in a suitable form, the next step is application of differentiation rules such as:

• Product rule: This is utilized where applicable to functions represented as multiplicative terms.
• Quotient rule: Useful for functions expressed as fractions.
• Chain rule: Especially pivotal when differentiating composite functions like our exercise.

Moreover, constants such as \(\ln(6)\) present in multiplied terms can be factored out, simplifying differentiation by treating them as constants.

Finally, simplifying the results remains vital. Take our derivative \(\frac{5}{(5x+1) \ln(6)}\). Ensure that any expression is simplified as much as possible, removing common factors where applicable. This approach guarantees not only accuracy but efficiency in computation.