Derivative of natural logs

Thread Starter

Distort10n

Joined Dec 25, 2006
429
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?
 

Attachments

Papabravo

Joined Feb 24, 2006
21,225
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.
 

Thread Starter

Distort10n

Joined Dec 25, 2006
429
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?
 

Attachments

Salgat

Joined Dec 23, 2006
218
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.
 

n9352527

Joined Oct 14, 2005
1,198
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.
 

Thread Starter

Distort10n

Joined Dec 25, 2006
429
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.
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.
 

n9352527

Joined Oct 14, 2005
1,198
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
 
Top