Question

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

Answer 1

The answer is `=1/e`

Explanation:

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`