# Derivative of natural logs

Discussion in 'Math' started by Distort10n, Jun 18, 2007.

1. ### Distort10n Thread Starter Active Member

Dec 25, 2006
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I need an extra pair of eyes for this problem. I am stuck on the algebra portion of it as silly as that is. I tried this problem for over an hour last night, and decided to ask for your opinion. I stopped halfway through at the point where I get lost. The answer is at the top, but for the life of me, I do not see how you get from point A to point B.

Thoughts?

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2. ### Papabravo Expert

Feb 24, 2006
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I think you made a mistake when you wrote the derivative of the log functions. Go back and review that process. Then take a look at the chain rule.

3. ### Distort10n Thread Starter Active Member

Dec 25, 2006
429
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Here is my updated work. I do not think there is a mistake in the log function. I had several people look at it so far, and it just seems to be a painful algebraic exercise.

The final answer that I get, the back of the book, and the TI-89 answer are equivalent when you plug in values. I used value of (2).

Any other thoughts?

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4. ### Salgat Active Member

Dec 23, 2006
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After testing your original solution, the book's solution, and Maple's solution, they all are equal and correct, so no mistakes that I know of, just a case of nasty simplification. If you find the solution for this I'm personally curious, as I am stuck too. Unfortunately Maple won't simplify as far as your book does.

5. ### n9352527 AAC Fanatic!

Oct 14, 2005
1,198
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Your final solution, if you divided all the terms with X^2 + 1, and then divided the resulting terms with sqrt(X^2 + 1) + 1, you would arrive at the solution given by the book. So, yes, your solution is correct.

6. ### Distort10n Thread Starter Active Member

Dec 25, 2006
429
1
I still do not see it. Maybe because it is midnight. Dividing all terms by x^2+1 will yield:

Numerator: sqrt(x^2 + 1) + 1

Denominator: x[(sqrt(x^2 + 1)/(X^2 + 1)) + 1]

From here, I do not see how dividing all terms by (sqrt(x^2 + 1) + 1) will make it simpler.

7. ### n9352527 AAC Fanatic!

Oct 14, 2005
1,198
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Yes, I skipped a step there. Sorry.

The denominator can also be represented by:

x[sqrt(X^2 + 1) + 1][1/sqrt(X^2 + 1)]

Now, dividing both denominator and numerator by sqrt(X^2 + 1) + 1, would result in:

numerator: 1
denominator: x/sqrt(X^2 + 1)

Taking the sqrt term up to numerator, or equal to multiplying both numerator and denominator by sqrt(X^2 + 1), the final result is:

sqrt(X^2 + 1)/x