Given $y= lnx/e^x$ how do you find find f'(1)?

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Answer 1 · 2016-10-30T08:51:47

The answer is $=1/e$

The function is a quotient of 2 derivable functions $u/v$
And the derivative is $(u/v)'=(u'v-uv')/v^2$

here $u=lnx$ $=>$ $u'=1/x$
and $v=e^x$ $=>$ $v'=e^x$

So $f'(x)=(1/x*e^x-e^xlnx)/(e^x)^2=(e^x(1/x-lnx))/(e^x)^2$
$=(1/x-lnx)/e^x$
so $f'(1)=(1-0)/e=1/e$

Given y= lnx/e^xy= lnx/e^x how do you find find f'(1)?