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OpenStudy (nincompoop):
try solving
OpenStudy (anonymous):
I attempted logarithmic differentiation but did not know how to do it with all the subtracting
OpenStudy (nincompoop):
?
OpenStudy (nincompoop):
show me your work
OpenStudy (anonymous):
all I did was take the natural log of everything and subtracted
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OpenStudy (nincompoop):
why not make it simpler by using quotient and chain rules?
OpenStudy (nincompoop):
natural log of everything?
OpenStudy (anonymous):
because isn't the natural log easier and what would be the u in the chain rule?
OpenStudy (nincompoop):
u = -x
OpenStudy (nincompoop):
isn't it neat?
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OpenStudy (anonymous):
no I don't get it
OpenStudy (nincompoop):
set up as a quotient rule first
OpenStudy (anonymous):
and then to the u chain rule afterwards?
OpenStudy (nincompoop):
ye
OpenStudy (nincompoop):
are you doing it or what?
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OpenStudy (anonymous):
yeah I doing it imam see how it comes out
OpenStudy (anonymous):
the top cancels out
OpenStudy (anonymous):
can I get some assitance
OpenStudy (anonymous):
does it help to know that this is \(\tanh(x)\) or do you not use hyperbolic functions?
OpenStudy (anonymous):
yeah I know what it is , is that needed
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OpenStudy (anonymous):
or you can multiply top and bottom by \(e^x\) and get
\[\frac{e^{2x}-1}{e^{2x}+1}\] if that makes it easier to use the quotient rule
there is really no way around the quotient rule here
OpenStudy (anonymous):
okay so its just quotient rule? and messy work? could I take the natural log of everything