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335833583358
>> No. 3358 Anonymous
16th September 2013
Monday 4:46 am
3358 spacer
(log x)' = 1/x
(log 2x)' = 1/x
(log 3x)' = 1/x
(log kx)' = 1/x
Expand all images.
>> No. 3359 Anonymous
16th September 2013
Monday 11:03 am
3359 spacer
Actually, you can explain it quite easily. I trust you're familiar with the chain rule?
>> No. 3360 Anonymous
16th September 2013
Monday 2:41 pm
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>>3359

Go on then
>> No. 3361 Anonymous
16th September 2013
Monday 3:47 pm
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>>3360
log(kx) is a function containing another function, so we use the chain rule to differentiate it.

Chain rule:
f(g(x))` = g`(x) f`(g(x))

kx is our g`(x) in this case so the differential becomes

d(log(kx))/dx = d(kx)/dx * d(log(g(x))/dx

= k * 1/kx = k/kx = 1/x

You sometimes see it written with us and vs instead of fs and gs, especially when using the d/dx notation rather than primes. The way I was taught is that you differentiate inside the brackets first and differentiate everything but the brackets and then just multiply them together.
>> No. 3362 Anonymous
16th September 2013
Monday 4:04 pm
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>>3358

Oh yeah?

(logkx)' = (logx + logk)' = 1/x + 0
>> No. 3363 Anonymous
16th September 2013
Monday 4:05 pm
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>>3362
Or that way too.
>> No. 3364 Anonymous
16th September 2013
Monday 4:06 pm
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>>3362

Conversely,

∫dx/x = logx + c = log(kx)
>> No. 3365 Anonymous
16th September 2013
Monday 4:10 pm
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It's interesting. Multiplying x by a constant only has the effect of shifting the graph line up or down so the slope stays the same.

http://www.fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJsb2coeCkiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjAsImVxIjoibG9nKDEweCkiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjAsImVxIjoibG9nKDEwMHgpIiwiY29sb3IiOiIjMDAwMDAwIn0seyJ0eXBlIjoxMDAwfV0-
>> No. 3366 Anonymous
16th September 2013
Monday 6:50 pm
3366 spacer
>>3365
That's how logarithms work. log kx = log k + log x. If k is constant, then so is log k.
>> No. 3367 Anonymous
16th September 2013
Monday 8:04 pm
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You_don't_say.png
336733673367
>>3366

(A good day to you Sir!)

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