anonymous
  • anonymous
does y' of ln(4^s) = 1/4^s or s/4 ??
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
\(\large ln(4^s) = s \cdot ln(4)\)
anonymous
  • anonymous
so it would be s/4 then.
anonymous
  • anonymous
ln(4) is a constant

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anonymous
  • anonymous
yes but the derivative of ln=1/x
anonymous
  • anonymous
If you are finding \(\large y'=\frac{d}{ds}[ ln(4^s)]\) \(\large y'=\frac{d}{ds}[ ln(4)s] =\frac{d}{ds}[c\cdot s], \space c=ln(4)\)
anonymous
  • anonymous
The derivative of ln x is 1/x, but ln 4 is a constant, and the derivative of a constant is zero.
anonymous
  • anonymous
But here you are taking the derivative of a constant times a variable.
anonymous
  • anonymous
what is c*s ???
anonymous
  • anonymous
im confused by your explanation.
TuringTest
  • TuringTest
the derivative is with respect to s what is d/ds(s) ?
anonymous
  • anonymous
umm s'?
TuringTest
  • TuringTest
no, what is d/dx(x) ?
anonymous
  • anonymous
s' would be if s was a function of some other variable.
anonymous
  • anonymous
=1
TuringTest
  • TuringTest
yes, so we just changed the name of s to x ...so d/ds(s)=?
anonymous
  • anonymous
=1 !
TuringTest
  • TuringTest
yes :) now what if we multiply by a constant that we call 'c' ? then d/ds(cs)=?
anonymous
  • anonymous
c
TuringTest
  • TuringTest
yes now for your problem we can write ln(4^s)=s*ln4 ln4 is a constant, so d/ds(ln(4^s))=?
anonymous
  • anonymous
ln4?
TuringTest
  • TuringTest
yes
anonymous
  • anonymous
so thats the answer?
TuringTest
  • TuringTest
yep
anonymous
  • anonymous
Thanks for the backup, Turing!
TuringTest
  • TuringTest
No prob!

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