March 18th, 2017, 06:56 PM  #1 
Newbie Joined: Nov 2016 From: USA Posts: 26 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 =(x1)/x Last edited by Seventy7; March 18th, 2017 at 07:04 PM. 
March 18th, 2017, 07:07 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,640 Thanks: 959 Math Focus: Elementary mathematics and beyond 
$\ln(1)=0$ it's a constant anyway, so the derivative would be zero (for any number).

March 18th, 2017, 07:13 PM  #3 
Newbie Joined: Nov 2016 From: USA Posts: 26 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! 
March 18th, 2017, 08:01 PM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,658 Thanks: 651 Math Focus: Wibbly wobbly timeywimey stuff.  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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