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Mathematics 8 Online
OpenStudy (anonymous):

Derivative of e^(lnx)^2

OpenStudy (anonymous):

Use the chain rule twice.

OpenStudy (anonymous):

how do u do it

OpenStudy (anonymous):

derivative of e^u = e^u * du/dx What is the u in this case?

OpenStudy (anonymous):

would it be (lnx)^2

OpenStudy (anonymous):

right. what is the derivative of u^2?

OpenStudy (anonymous):

2u

OpenStudy (anonymous):

right 2u * du/dx

OpenStudy (anonymous):

2lnx * ??

OpenStudy (anonymous):

its 1/x^2

OpenStudy (anonymous):

just the 1/x, the square is on the "outside"

OpenStudy (anonymous):

therefore its 2 over x

OpenStudy (anonymous):

put it all together. e^ln x ^2 * 2ln x * 1/x and clean it up.

OpenStudy (anonymous):

OK? Gotta go. Good luck with it.

OpenStudy (angela210793):

it's e^U

OpenStudy (angela210793):

\[=(2\log^2_{x} logx)/x\] I guess....

OpenStudy (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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