How Do You Differentiate a Natural Logarithm Function Like This?

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Discussion Overview

The discussion revolves around the differentiation of a natural logarithm function, specifically the function $$f(x)=\ln\left[{\frac{(2x+3)(x+6)^5}{(1-2x)^3}}\right].$ The participants explore various steps involved in the differentiation process, including expansion and the application of logarithmic properties.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant proposes to expand the logarithm using properties of logarithms, breaking it down into simpler components: $$f(x)=\ln\left({2x+3}\right) +5\ln\left({x+6}\right) -3\ln\left({1-2x}\right).$$
  • Another participant suggests using the derivative of the logarithm function, stating that $$\dfrac{d\ln(f(x))}{dx}=\dfrac{f'(x)}{f(x)}.$$
  • A participant questions the thread title "Ln integral," pointing out that the problem is about finding a derivative, not an integral.
  • One participant provides a derivative result for the function, stating $$f'(x)=\frac{3\left(4{x}^{2}-16x-45 \right)}{(x+6)(2x-1)(2x+3)}$$ and reiterates the expansion of the logarithm.
  • Another participant emphasizes the step-by-step differentiation of each term, indicating that for each term, different functions are used for \(f(x)\) in the derivative calculation.

Areas of Agreement / Disagreement

Participants express differing views on the title of the thread and the focus of the discussion. While there is some agreement on the steps to differentiate the function, the discussion remains unresolved regarding the appropriateness of the title and the clarity of the problem being addressed.

Contextual Notes

Some participants' contributions include assumptions about the differentiation process and the use of logarithmic properties, but these assumptions are not universally accepted or clarified. The discussion does not resolve the potential confusion regarding the title and its relevance to the differentiation task.

karush
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$\large{242.7.3.83}$

Differentiate
$$\displaystyle
f(x)=\ln\left[{\frac{(2x+3)(x+6)^5}{(1-2x)^3}}\right]$$
Assume first step is expansion..
$$f(x)=\ln\left({2x+3}\right)
+5\ln\left({x+6}\right)
-3\ln\left({1-2x}\right)$$
 
Last edited:
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Re: Ln integral. 242.7.3.83

Yes. Now use

$$\dfrac{d\ln(f(x))}{dx}=\dfrac{f'(x)}{f(x)}$$
 
Re: Ln integral. 242.7.3.83

Why was this titled "Ln integral" when the problem asks for a derivative?
 
$\large{242.7.3.83}$

Differentiate
$$\displaystyle
f(x)=\ln\left[{\frac{(2x+3)(x+6)^5}{(1-2x)^3}}\right] \ \ \ \ \
f'(x)=\frac{3\left(4{x}^{2}-16x-45 \right)}{(x+6)(2x-1)(2x+3)}$$
expansion..
$$f(x)=\ln\left({2x+3}\right)
+5\ln\left({x+6}\right)
-3\ln\left({1-2x}\right) $$
then
$$\dfrac{f'(x)}{f(x)}
= \frac{2}{2x+3}+\frac{5}{x+6}-\frac{6}{2x-1}$$$$$$
 
Last edited:
karush said:
$\large{242.7.3.83}$

Differentiate
$$\displaystyle
f(x)=\ln\left[{\frac{(2x+3)(x+6)^5}{(1-2x)^3}}\right] \ \ \ \ \
f'(x)=\frac{3\left(4{x}^{2}-16x-45 \right)}{(x+6)(2x-1)(2x+3)}$$
expansion..
$$f(x)=\ln\left({2x+3}\right)
+5\ln\left({x+6}\right)
-3\ln\left({1-2x}\right) $$
then
$$\dfrac{d\ln(2x+3)}{dx}=\dfrac{f'(x)}{f(x)}$$
In this case, since you are doing this one term at a time, f(x) = 2x + 3. Then for the second term f(x) = x + 6 and for the third f(x) = 1 - 2x

-Dan
 

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