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Chapter 4: Problem 34

Differentiate. $$y=\ln [(x+1)(2 x+1)(3 x+1)]$$

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} \)

Step by step solution

01

Apply the Logarithm Property

Use the property of logarithms that states \(\text{ln}(a \times b \times c) = \text{ln}(a) + \text{ln}(b) + \text{ln}(c)\). This turns the equation into: \(\text{ln}[(x+1)(2x+1)(3x+1)] = \text{ln}(x+1) + \text{ln}(2x+1) + \text{ln}(3x+1)\)
02

Differentiate Each Term

Differentiate each term using the chain rule. We differentiate \(\text{ln}(u)\) as \(\frac{1}{u} \frac{du}{dx}\). The derivatives are: \(\frac{d}{dx}[\text{ln}(x+1)] = \frac{1}{x+1}\), \(\frac{d}{dx}[\text{ln}(2x+1)] = \frac{1}{2x+1} \times 2 = \frac{2}{2x+1}\), \(\frac{d}{dx}[\text{ln}(3x+1)] = \frac{1}{3x+1} \times 3 = \frac{3}{3x+1}\)
03

Combine the Derivatives

Add the derivatives together: \(\frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1}\). This represents the derivative of the original function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

logarithmic differentiation
Logarithmic differentiation is a useful technique for simplifying the differentiation process.
It's particularly helpful for functions that are products, quotients, or powers of other functions.
The method involves taking the natural logarithm of both sides of an equation.
This often simplifies the differentiation process, especially when combined with the properties of logarithms.
For instance, in our exercise, taking the natural logarithm of \( \text{ln}[(x+1)(2x+1)(3x+1)] \) helps break down a complex multiplication into simpler terms that are easier to differentiate.
chain rule
The chain rule is essential in differentiation, particularly when dealing with composite functions.
The rule states that if you have a compound function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is \( f'(g(x)) \times g'(x) \).
In our exercise, each term \( \text{ln}(x+1), \text{ln}(2x+1), \text{ln}(3x+1) \) can be seen as a composite function.
For example, to differentiate \( \text{ln}(2x+1) \), you first consider its outer function (the logarithm) and then its inner function ( \( 2x+1 \)).
This results in the derivative being \( \frac{1}{2x+1} \times 2 \), simplifying to \( \frac{2}{2x+1} \).
logarithm properties
Understanding the properties of logarithms can make the differentiation process a lot smoother.
Some key properties are:
  • \( \text{ln}(a \times b) = \text{ln}(a) + \text{ln}(b) \)
  • \( \text{ln}(a^b) = b \times \text{ln}(a) \)
  • \( \text{ln}(e) = 1 \)

In our exercise, we use the property \( \text{ln}(a \times b \times c) = \text{ln}(a) + \text{ln}(b) + \text{ln}(c) \).
This allows us to break down the complex product \( (x+1)(2x+1)(3x+1) \) into manageable chunks.
Once separated, each chunk's differentiation becomes more straightforward.

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