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March 18th, 2017, 05:56 PM   #1
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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


Last edited by Seventy7; March 18th, 2017 at 06:04 PM.
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March 18th, 2017, 06:07 PM   #2
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$\ln(1)=0$ it's a constant anyway, so the derivative would be zero (for any number).
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March 18th, 2017, 06:13 PM   #3
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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.

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March 18th, 2017, 07:01 PM   #4
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Originally Posted by Seventy7 View Post
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!

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