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OpenStudy (anonymous):
ln(xy) = chain rule and then a product rule.
OpenStudy (anonymous):
how about the left side?
OpenStudy (anonymous):
That is the left side
OpenStudy (anonymous):
OH OK
OpenStudy (anonymous):
:)
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OpenStudy (anonymous):
^^
OpenStudy (anonymous):
Thank you
OpenStudy (anonymous):
yw
OpenStudy (anonymous):
hint ---> ln(xy) = ln(x) + ln(y) = ln(x) + e^ln(ln(y)) = e^ln(ln(e^ln(y)))) + ln(x)
hope this helps
OpenStudy (anonymous):
I still don't get the right answer
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zepdrix (zepdrix):
Hey mr koala c:
Figure this out yet?
OpenStudy (anonymous):
not yet
zepdrix (zepdrix):
\[\Large\sf \ln(xy)=x+y\]Derivative of the left side,\[\Large\sf \frac{1}{xy}\color{royalblue}{(xy)'}\]Log gives us one over the stuff right?
Then we have to chain rule, multiplying by the derivative of the inner function.
So we need to differentiate this blue part.
Product rule.\[\Large\sf \frac{1}{xy}\left(\color{royalblue}{x'}y+x\color{royalblue}{y'}\right)\]Make sense so far?
OpenStudy (anonymous):
Yay zep :3, always making it look so fancy and nice!
OpenStudy (anonymous):
Mr koala i hope you understand how to do it now, with zepdrix's awesome explanation :d
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