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

Answer 1

Charles Anctil

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#

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Answer 2

Cameron Alexander

To find ( f'(1) ) for ( y = \frac{\ln x}{e^x} ), first differentiate the function to find ( f'(x) ), then evaluate it at ( x = 1 ).

[ f'(x) = \frac{\frac{1}{x} \cdot e^x - \ln x \cdot e^x}{(e^x)^2} ]

Now, substitute ( x = 1 ) into ( f'(x) ) to find ( f'(1) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.