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

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- May 5th 2007, 11:11 AMbeckyquotent 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 AMJhevon
- May 5th 2007, 11:31 AMJhevon
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 PMbeckyInx/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 PMJhevon
- May 5th 2007, 12:41 PMqbkr21Re:
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 PMqbkr21Re:
Re: