anonymous
  • anonymous
Given c is some constant Why is the derivative for d/dx Log [c x] = 1/x ? And not c Log [ c x] ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I mean what kind of logical thought process should I have to lock that in ?
Australopithecus
  • Australopithecus
Assuming it is log based 10 \[\frac{d}{dx} \log_a(u(x)) = \frac{u'(x)}{\ln(a)*u(x)}\]
anonymous
  • anonymous
I get that d/dx Log [x] = 1/x and normally c = 0 but then if I think chain rule .. I cook up .. Log [c x] (c)(x) (c) by constant multiple c Log[ c x ] But when I put Log [c x] into mathematica, it spits out 1/x ???

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Australopithecus
  • Australopithecus
its ln(x)?
anonymous
  • anonymous
oh assuming oh assuming base E
Australopithecus
  • Australopithecus
this formula still works for ln(x) because ln(e) = 1
phi
  • phi
see https://www.khanacademy.org/math/differential-calculus/taking-derivatives/der_common_functions/v/proof-d-dx-ln-x-1-x
Australopithecus
  • Australopithecus
The best way two approach derivatives is to split a function into smaller functions. For example for a function f(x) = ln(ax^2) notice f(x) is just two functions g(x) = ln(x) m(x) =ax^2 differentiate m(x) an g(x), we get g(x) = ln(x) g'(x) = 1/x m(x) =ax^2 m'(x) = 2ax now plug them into the chain rule formula f'(x) = g'(m(x))*m'(x) so pluging it in f'(x) = (1/(ax^2))(2ax)
Australopithecus
  • Australopithecus
the constant cancels out
Australopithecus
  • Australopithecus
For the derivative
phi
  • phi
\[ \frac{d}{du} \ln u = \frac{1}{u} du \] if u = cx then \[ \frac{d}{dx} \ln c x = \frac{1}{cx} \frac{d}{dx} cx = \frac{c}{cx} = \frac{1}{x}\]
Australopithecus
  • Australopithecus
remember the derivative of ax is a
Australopithecus
  • Australopithecus
d/dx ax = ax^0 = a(1) = a
Australopithecus
  • Australopithecus
and by the chain rule formula g'(m(x))*m'(x) you should see that a wont leave the function and since it is a log it will cancel out
anonymous
  • anonymous
ahh.. gotcha. *Click*
anonymous
  • anonymous
so my mistake was not taking the derivative for both sides, when applying the chain rule. Ln [ c x] u=cx Ln [u] (cx) 1/u c taking derivatives of both terms expand u back again c/cx 1/x simplified.
Australopithecus
  • Australopithecus
essentially
anonymous
  • anonymous
probably not the best way to notate it.. but as a general identity I can say then Ln [c x] = 1/x by chain rule?
anonymous
  • anonymous
d/dx Ln[c x] = 1/x by chain rule...
Australopithecus
  • Australopithecus
yes
Australopithecus
  • Australopithecus
Look at the two graphs, note that a constant doesnt change the slope
Australopithecus
  • Australopithecus
or rather a constant inside a logarithm doesn't effect the slope o
Australopithecus
  • Australopithecus
It just shifts the function
perl
  • perl
*

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