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anonymous

  • one year ago

Given c is some constant Why is the derivative for d/dx Log [c x] = 1/x ? And not c Log [ c x] ?

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  1. anonymous
    • one year ago
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    I mean what kind of logical thought process should I have to lock that in ?

  2. Australopithecus
    • one year ago
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    Assuming it is log based 10 \[\frac{d}{dx} \log_a(u(x)) = \frac{u'(x)}{\ln(a)*u(x)}\]

  3. anonymous
    • one year ago
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    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 ???

  4. Australopithecus
    • one year ago
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    its ln(x)?

  5. anonymous
    • one year ago
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    oh assuming oh assuming base E

  6. Australopithecus
    • one year ago
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    this formula still works for ln(x) because ln(e) = 1

  7. Australopithecus
    • one year ago
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    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)

  8. Australopithecus
    • one year ago
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    the constant cancels out

  9. Australopithecus
    • one year ago
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    For the derivative

  10. phi
    • one year ago
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    \[ \frac{d}{du} \ln u = \frac{1}{u} du \] if u = cx then \[ \frac{d}{dx} \ln c x = \frac{1}{cx} \frac{d}{dx} cx = \frac{c}{cx} = \frac{1}{x}\]

  11. Australopithecus
    • one year ago
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    remember the derivative of ax is a

  12. Australopithecus
    • one year ago
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    d/dx ax = ax^0 = a(1) = a

  13. Australopithecus
    • one year ago
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    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

  14. anonymous
    • one year ago
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    ahh.. gotcha. *Click*

  15. anonymous
    • one year ago
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    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.

  16. Australopithecus
    • one year ago
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    essentially

  17. anonymous
    • one year ago
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    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?

  18. anonymous
    • one year ago
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    d/dx Ln[c x] = 1/x by chain rule...

  19. Australopithecus
    • one year ago
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    yes

  20. Australopithecus
    • one year ago
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    Look at the two graphs, note that a constant doesnt change the slope

  21. Australopithecus
    • one year ago
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    or rather a constant inside a logarithm doesn't effect the slope o

  22. Australopithecus
    • one year ago
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    It just shifts the function

  23. perl
    • one year ago
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    *

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