If y=xe^(-1/x) is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x)
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Originally Posted by spazzyskylar If y=xe^(-1/x) is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x) You said d/dx(-1/x) = -(x-1) / x^2 Are you sure about this?
Originally Posted by spazzyskylar If y=xe^(-1/x) is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x) ? note that the derivative of is
Not really, should it be -(x+1)/x^2
Not quite! if
Last edited by Archie Meade; Jan 3rd 2010 at 03:07 PM.
Originally Posted by spazzyskylar If y=xe^(-1/x) is y'= xe^(-1/x)*(-(x-1)/x^2)+e^(-1/x) skeeter gave the derivative of the exponent so multiply due to the chain rule Now use the product rule y' = u'v + uv' Spoiler:
I got it now
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