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Mathematics 9 Online
OpenStudy (anonymous):

What's the derivative of (1-2xe^x)/x

OpenStudy (aaronq):

(1-2xe^x)/x = (1-2xe^x)(x^-1) use the product rule

OpenStudy (anonymous):

I have to distribute the x^-1 first right? or can I just take the derivative of them now? How would i take the derivative of 2xe^x?

OpenStudy (aaronq):

no don't distribute. il show you an example: d/dx (xe^x) Using product rule d/dx (uv) = u'v + uv' u = x u' = 1 v = e^x v' = e^x d/dx (xe^x) = e^x + xe^x = e^x(1+x)

OpenStudy (anonymous):

I'm lost... In class we never went over a product rule so Is it possible to solve via (1/x)-[(2xe^x)/x] and then cancel x to get 1/x - (2e^x) using the power rule and knowing that the derivative of e^x is e^x I could conclude that the derivative would be [-1/(x^2)]-2e^x?

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