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bmelyk

  • 3 years ago

does y' of ln(4^s) = 1/4^s or s/4 ??

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  1. CliffSedge
    • 3 years ago
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    \(\large ln(4^s) = s \cdot ln(4)\)

  2. bmelyk
    • 3 years ago
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    so it would be s/4 then.

  3. CliffSedge
    • 3 years ago
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    ln(4) is a constant

  4. bmelyk
    • 3 years ago
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    yes but the derivative of ln=1/x

  5. CliffSedge
    • 3 years ago
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    If you are finding \(\large y'=\frac{d}{ds}[ ln(4^s)]\) \(\large y'=\frac{d}{ds}[ ln(4)s] =\frac{d}{ds}[c\cdot s], \space c=ln(4)\)

  6. CliffSedge
    • 3 years ago
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    The derivative of ln x is 1/x, but ln 4 is a constant, and the derivative of a constant is zero.

  7. CliffSedge
    • 3 years ago
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    But here you are taking the derivative of a constant times a variable.

  8. bmelyk
    • 3 years ago
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    what is c*s ???

  9. bmelyk
    • 3 years ago
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    im confused by your explanation.

  10. TuringTest
    • 3 years ago
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    the derivative is with respect to s what is d/ds(s) ?

  11. bmelyk
    • 3 years ago
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    umm s'?

  12. TuringTest
    • 3 years ago
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    no, what is d/dx(x) ?

  13. CliffSedge
    • 3 years ago
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    s' would be if s was a function of some other variable.

  14. bmelyk
    • 3 years ago
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    =1

  15. TuringTest
    • 3 years ago
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    yes, so we just changed the name of s to x ...so d/ds(s)=?

  16. bmelyk
    • 3 years ago
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    =1 !

  17. TuringTest
    • 3 years ago
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    yes :) now what if we multiply by a constant that we call 'c' ? then d/ds(cs)=?

  18. bmelyk
    • 3 years ago
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    c

  19. TuringTest
    • 3 years ago
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    yes now for your problem we can write ln(4^s)=s*ln4 ln4 is a constant, so d/ds(ln(4^s))=?

  20. bmelyk
    • 3 years ago
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    ln4?

  21. TuringTest
    • 3 years ago
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    yes

  22. bmelyk
    • 3 years ago
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    so thats the answer?

  23. TuringTest
    • 3 years ago
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    yep

  24. CliffSedge
    • 3 years ago
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    Thanks for the backup, Turing!

  25. TuringTest
    • 3 years ago
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    No prob!

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