# quotent rule for natural logs

• May 5th 2007, 11:11 AM
becky
quotent rule for natural logs
f(x) = (lnx)x
f'(x) = x * (lnx)' - (lnx) * x'/(x)^2
f'(x) = x*(1/x) - (lnx)*1/(x)^2
f'(x) = 1 - lnx/(x)^2

I hope this is correct
• May 5th 2007, 11:26 AM
Jhevon
Quote:

Originally Posted by becky
f(x) = (lnx)x
f'(x) = x * (lnx)' - (lnx) * x'/(x)^2
f'(x) = x*(1/x) - (lnx)*1/(x)^2
f'(x) = 1 - lnx/(x)^2

I hope this is correct

it's not correct i'm afraid. you would use the product rule here, since you don't have a quotient (or did you mean to type a slash after the lnx but forgot?)

y = (ln)x
=> y' = lnx + x(1/x) = lnx + 1
• May 5th 2007, 11:31 AM
Jhevon
Was this what you meant? If so, you were correct (though you need to have parenthesis in your answer to show that this is the same thing)
• May 5th 2007, 12:05 PM
becky
Inx/x
Yes, that's what the problem...
Glad you caught my mistake in writing the problem, I probably wouldn't have and would really be confused right now...
Yeah, I got it right!
• May 5th 2007, 12:20 PM
Jhevon
Quote:

Originally Posted by becky
Yes, that's what the problem...
Glad you caught my mistake in writing the problem, I probably wouldn't have and would really be confused right now...
Yeah, I got it right!

Good Job:D

so you understand it well now?

why not try a few more problems and post your solutions
• May 5th 2007, 12:41 PM
qbkr21
Re:
Jhevon's is correct, but I think he will agree that unless you are told to do so using the Quotient rule isn't the easiest way to solve this problem...
• May 5th 2007, 12:47 PM
qbkr21
Re:
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