Problem 22 - Calculus and its applications

Question

Differentiate. $$ y=\log _{8}\left(x^{3}+x\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{3x^2 + 1}{\text{ln}(8) \cdot (x^3 + x)}.\)

Step 1: Apply the Logarithmic Differentiation Formula

To differentiate the function, use the change of base formula for logarithms. Rewrite the function as \[y = rac{ ext{ln}(x^3 + x)}{ ext{ln}(8)}.\]This allows us to differentiate using the natural logarithm.

Step 2: Differentiate the Natural Logarithm

Differentiate \( ext{ln}(x^3 + x)\) using the chain rule. The derivative of \( ext{ln}(u)\) is \(\frac{1}{u}\cdot \frac{du}{dx}\), where \(u = x^3 + x\). Therefore, \[\frac{d}{dx}[\text{ln}(x^3 + x)] = \frac{1}{x^3 + x} \cdot (3x^2 + 1).\]

Step 3: Combine Results

Now, divide the result from Step 2 by \(\text{ln}(8)\) as per the rewritten formula: \[\frac{d}{dx}[y] = \frac{1}{\text{ln}(8)} \cdot \left( \frac{3x^2 + 1}{x^3 + x} \right).\]Thus, the derivative is \[\frac{3x^2 + 1}{\text{ln}(8) \cdot (x^3 + x)}.\]

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a technique that is particularly useful when dealing with complex functions that are products, quotients, or in the form of an exponent. This method can simplify the differentiation process, especially when working with logarithmic functions.
To start, we often use the property of logarithms to transform a given function into a form that is easier to differentiate. In this exercise, the original function is a logarithm with a base of 8: \(y=\log_{8}(x^{3}+x)\).
To apply logarithmic differentiation, we first use the change of base formula, which expresses the logarithm with any base in terms of the natural logarithm (ln). The formula is given by:

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

Using this formula, we rewrite the function as \(y = \frac{\text{ln}(x^3 + x)}{\text{ln}(8)}\). This step allows us to differentiate the function using the properties of the natural logarithm.

Chain Rule

The Chain Rule is a fundamental rule in calculus for differentiating compositions of functions. In simpler terms, it helps us find the derivative when we have a function inside another function, which is often denoted as \(f(g(x))\).
For this exercise, after rewriting the original function using the natural logarithm, we turn to the component \(\text{ln}(x^3 + x)\). This is where the Chain Rule becomes relevant.
The Chain Rule formula is:

• If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\)

To apply the Chain Rule here, identify:

• The outer function: \(\text{ln}(u)\)
• The inner function: \(u = x^3 + x\)

The derivative of \(\text{ln}(u)\) with respect to \(u\) is \(\frac{1}{u}\), and the derivative of \(u = x^3 + x\) is \(3x^2 + 1\). Thus, applying the Chain Rule gives us: \(\frac{d}{dx}[\text{ln}(x^3 + x)] = \frac{1}{x^3 + x} \cdot (3x^2 + 1)\). This result is crucial for completing the differentiation process.

Natural Logarithm

Natural logarithms, indicated by \(\text{ln}\), are logarithms whose base is the irrational number \(e\), approximately equal to 2.71828. This is the base most widely used in calculus due to its unique mathematical properties.
The natural logarithm has several important rules that make differentiation straightforward, particularly the rule that the derivative of \(\text{ln}(u)\) with respect to \(u\) is \(\frac{1}{u}\). This is key when logarithmic differentiation is applied, as seen in the exercise.
In the problem at hand, once we rewrote \(y = \log_{8}(x^{3}+x)\) as \(y = \frac{\text{ln}(x^3 + x)}{\text{ln}(8)}\), the focus shifted to differentiating \(\text{ln}(x^3 + x)\). Thanks to the use of natural logarithms, we can directly apply its derivative property:
Here's the step-by-step:

• Take the derivative of \(\text{ln}(x^3 + x)\)
• Apply the result: \(\frac{1}{x^3 + x}\) and multiply by the derivative of \(x^3 + x\), which is \(3x^2 + 1\)

This way, natural logarithms simplify handling otherwise complex differentiation tasks, showing their significance in calculus and in exercises like this one.