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



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CliffSedge
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\(\large ln(4^s) = s \cdot ln(4)\)

bmelyk
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so it would be s/4 then.

CliffSedge
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ln(4) is a constant

bmelyk
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yes but the derivative of ln=1/x

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

CliffSedge
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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.

CliffSedge
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But here you are taking the derivative of a constant times a variable.

bmelyk
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what is c*s ???

bmelyk
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im confused by your explanation.

TuringTest
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the derivative is with respect to s
what is d/ds(s) ?

bmelyk
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umm s'?

TuringTest
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no, what is d/dx(x) ?

CliffSedge
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s' would be if s was a function of some other variable.

bmelyk
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=1

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

bmelyk
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=1 !

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

bmelyk
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c

TuringTest
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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))=?

bmelyk
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ln4?

TuringTest
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yes

bmelyk
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so thats the answer?

TuringTest
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yep

CliffSedge
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Thanks for the backup, Turing!

TuringTest
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No prob!