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

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Homework Statement



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.

Homework Equations



Derivative of ln(x) = 1/x


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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both the methods are incorrect
d/dx(log 2x)=(1/2x)d/dx(2x)
=1/x
 
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Arshad_Physic said:
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.
 
Thanks Slider and Monty! :)
 
(d(ln 2x)/ dx) / (d(2x)/ dx) = 2/2x/2 = 1/2x
 
bobn said:
(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.
 
ohh sorry I calculatd, derivative of ln2x wrt to 2x.
 
1/2x
 
fan_103 said:
1/2x
try reading the other posts... d(ln2x)/dx = 1/x
 
  • #10
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'
 
  • #11
duke222 said:
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.
 
  • #12
I didn't see it mentioned but observe also you can apply the properties of logarithms:

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

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