COnfused: what is the derivative of ln(2x)?


by Arshad_Physic
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Arshad_Physic
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#1
Oct17-09, 04:38 AM
P: 51
1. The problem statement, all variables and given/known data

What is the derivative of ln(2x)?

I was just thinking about this, and I got 2 answers. I am in Calc 2 right now.

2. Relevant equations

Derivative of ln(x) = 1/x


3. The attempt at a solution

Since d/dx lna = (1/a)*(derivative of a)

Thus d/dx ln2x = (1/2x)*(2)

BUT

I can also do this, I think: d/dx ln2x = 2d/dx lnx = 2*1/x = 2/x

I am CONFUSED!! lol !:)

Please tell me which is the correct method! :)

Thanks! :)
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monty37
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#2
Oct17-09, 04:51 AM
P: 225
both the methods are incorrect
d/dx(log 2x)=(1/2x)d/dx(2x)
=1/x
slider142
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#3
Oct17-09, 06:35 AM
P: 878
Quote Quote by Arshad_Physic View Post

Since d/dx lna = (1/a)*(derivative of a)

Thus d/dx ln2x = (1/(2x))*(2)
This is correct. Note that ln(ax) = ln(a) + ln(x). Since ln(a) is a constant, the derivative is always 1/x, irrespective of 'a'. In geometric terms, 'a' simply moves the graph of the logarithm up or down; it does not change the shape of the graph.

BUT

I can also do this, I think: d/dx ln2x = 2d/dx lnx
This is wrong. The natural logarithm is not linear: you cannot pull the 2 out of the ln, irrespective of the derivative. ln(2x) is not 2ln(x) any more than cos(2x) = 2cos(x). It would be a good idea to review the definition and properties of logarithms.

Arshad_Physic
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#4
Oct17-09, 01:11 PM
P: 51

COnfused: what is the derivative of ln(2x)?


Thanks Slider and Monty!! :)
bobn
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#5
Oct18-09, 10:23 AM
P: 22
(d(ln 2x)/ dx) / (d(2x)/ dx) = 2/2x/2 = 1/2x
HallsofIvy
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#6
Oct18-09, 04:05 PM
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Quote Quote by bobn View Post
(d(ln 2x)/ dx) / (d(2x)/ dx) = 2/2x/2 = 1/2x
100% wrong! Go back and read the previous responses to this question. The derivative is 1/x.
bobn
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#7
Oct18-09, 09:58 PM
P: 22
ohh sorry I calculatd, derivative of ln2x wrt to 2x.
fan_103
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#8
Oct19-09, 09:06 AM
P: 24
1/2x
lanedance
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#9
Oct19-09, 09:21 AM
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Quote Quote by fan_103 View Post
1/2x
try reading the other posts... d(ln2x)/dx = 1/x
duke222
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#10
Jan18-10, 05:42 AM
P: 2
anti derivative of 1/x or x^-1 = ln (x) natural log of x =ln x +c so the derivative of c + ln (2x)dx=1/2x +C'
Mark44
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#11
Jan18-10, 12:42 PM
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Quote Quote by duke222 View Post
anti derivative of 1/x or x^-1 = ln (x) natural log of x =ln x +c so the derivative of c + ln (2x)dx=1/2x +C'
Wrong on two counts:
  1. d/dx(c) = 0 - not c'
  2. d/dx(ln(2x)) = 1/x - you are forgetting to use the chain rule.
jambaugh
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#12
Jan18-10, 02:08 PM
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I didn't see it mentioned but observe also you can apply the properties of logarithms:

[tex] d/dx \, \ln(2x) = d/dx\, [\ln(x) + \ln(2)] = 1/x + 0[/tex]


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