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 March 18th, 2017, 05:56 PM #1 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 Quick derivative question Find the derivative of ln(1/x). Rewrite as lnx^-1, which equals -lnx, whose derivative answer is -1/x. ------------------------------------------------------------------------------- Why won't this method produce the same answer for the same problem above?: ln(1/x) = d/dx[ln1- lnx]; law of logs = 1/1 - 1/x; since d/dx (ln x) = 1/x, and by that same formula d/dx (ln 1) = 1/1 ? = 1 - 1/x = x/x - 1/x; common denominator =(x-1)/x Last edited by Seventy7; March 18th, 2017 at 06:04 PM.
 March 18th, 2017, 06:07 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond $\ln(1)=0$ it's a constant anyway, so the derivative would be zero (for any number). Thanks from Seventy7
 March 18th, 2017, 06:13 PM #3 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 That's right. I got caught up in the formula d/dx lnx = 1/x and forgot that ln 1 evaluates to a constant, whose derivative is zero. Thanks! Thanks from greg1313
March 18th, 2017, 07:01 PM   #4
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Quote:
 Originally Posted by Seventy7 ln(1/x) = d/dx[ln1- lnx]; law of logs
Pet Peeve Warning!! Danger, Will Robinson! Danger!

The above equation is not correct. ln(1/x) is not equal to d/dx[ln(1) - ln(x)]. Now if you wrote ln(1/x) = ln(1) - ln(x) then you are okay.

This has the appearance of writing thoughts down as you have them. I've seen expressions of the form 2x + 1 = 2x + 1 - 1 = x = -1/2. It's stream of thought but is very sloppy. And if your grader can't figure it out you'll lose points on an exam.

Pet Peeve Over and Out!

-Dan

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