anonymous
  • anonymous
Derivative of e^(lnx)^2
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Use the chain rule twice.
anonymous
  • anonymous
how do u do it
anonymous
  • anonymous
derivative of e^u = e^u * du/dx What is the u in this case?

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anonymous
  • anonymous
would it be (lnx)^2
anonymous
  • anonymous
right. what is the derivative of u^2?
anonymous
  • anonymous
2u
anonymous
  • anonymous
right 2u * du/dx
anonymous
  • anonymous
2lnx * ??
anonymous
  • anonymous
its 1/x^2
anonymous
  • anonymous
just the 1/x, the square is on the "outside"
anonymous
  • anonymous
therefore its 2 over x
anonymous
  • anonymous
put it all together. e^ln x ^2 * 2ln x * 1/x and clean it up.
anonymous
  • anonymous
OK? Gotta go. Good luck with it.
angela210793
  • angela210793
it's e^U
angela210793
  • angela210793
\[=(2\log^2_{x} logx)/x\] I guess....
anonymous
  • anonymous
Mmmm.... not so sure. Try this: we know d/dx(e^x) = e^x So d/dx(e^(lnx)^2) = e^(lnx)^2 times d/dx(lnx)^2 = e^(lnx)^2 times 2lnx times d/dx(lnx) = e^(lnx)^2 times 2lnx times (1/x)

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