# differentiating and integrating ln xWatch

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#1
0
14 years ago
#2
(Original post by rachplatt)
Is this ln x? (i.e. not lnmx?)

Aitch
0
14 years ago
#3
(Original post by Aitch)
Is this ln x? (i.e. not lnmx?)

Aitch
Assuming yes!

int lnx dx

INTEGRATE BY PARTS

v=lnx so dv/dx = 1/x

du/dx = 1 so du=dx so u = x

so int lnx (*1) = xlnx - int x*(1/x) dx = xlnx - x

differential of lnx is 1/x

Aitch
0
#4
Thanks for your help...would ln 3x be 1/3x or 3/x?
0
14 years ago
#5
(Original post by rachplatt)
Thanks for your help...would ln 3x be 1/3x or 3/x?
It would be neither - it would be 1/x.

You can view this by either writing

ln(3x) = lnx + ln3

and ln3 is a constant which differentiates to zero.

Or you can apply the chain rule and get

3/(3x) = 1/x

Generally, ln(f(x)) differentiates to

f'(x)/f(x)

and it's a formula worth remembering for when you need to integrate something of that form.
0
14 years ago
#6
(Original post by RichE)
It would be neither - it would be 1/x.

You can view this by either writing

ln(3x) = lnx + ln3

and ln3 is a constant which differentiates to zero.

Or you can apply the chain rule and get

3/(3x) = 1/x

Generally, ln(f(x)) differentiates to

f'(x)/f(x)

and it's a formula worth remembering for when you need to integrate something of that form.
This is also interesting when you come back in the other direction, say with

Given dy/dx = 1/x, find y, given that when x=1, y = ln3.

y = lnx + c

given that when x=1, y = ln3

ln3 = ln1 + c
c = ln3

so y = lnx + ln3 = ln(3x) which is where you started.

Aitch
0
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