anonymous
  • anonymous
what's the derivative of y=ln square root of 1+ sin2x
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[y= \ln \sqrt{1+\sin2x}\]
amistre64
  • amistre64
logs pull out exponents ... ln(a^b) = b ln(a) that should at least simplify the process
amistre64
  • amistre64
take the "e" of both sides, that will undo the ln

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amistre64
  • amistre64
since the constant is trivial .... \[y= ln(x)\] \[e^y= e^{ln(x)}\] \[e^y= x\] \[y'e^y= x'\] \[y'= e^{-y}\]
amistre64
  • amistre64
i feel like if made an error ....
amistre64
  • amistre64
in this case, i was trying to signify x' as the derivative of the innards ... so x' not equal to 1. bad choice of variable name on my part :)
anonymous
  • anonymous
why you took 'e'? no need to cancel out the ln .. can't you just do this y=ln (1+sin2x)^1/2
amistre64
  • amistre64
yeah, i was thinking integration with that .... i blame these tired old eyes :/
amistre64
  • amistre64
good catch tho ... \[y= \ln \sqrt{1+\sin2x}\] \[y= \frac12\ln (1+\sin2x)\]
anonymous
  • anonymous
didn't get what you did..
amistre64
  • amistre64
i just made it less scary by pulling out the sqrt of it.
anonymous
  • anonymous
how it came 1/2 ln ...
amistre64
  • amistre64
\[\sqrt{n}=n^{1/2}\] \[log(a^b)=b~log(a)\]
amistre64
  • amistre64
these are algebraic propeties
amistre64
  • amistre64
and the derivaitive of a log is:\[D[log(u)]=\frac{u'}{u}\]
anonymous
  • anonymous
ohhhh this point:) thaanks:D
anonymous
  • anonymous
you made it easy sir
amistre64
  • amistre64
:) i do get there in the end
anonymous
  • anonymous
hats off to you

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