How do you find the slope of a tangent line to the graph of the function #y=e^- x/(x+1)#, at x=1?

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Answer 1 · 2017-06-10T02:58:46

#-3/(4e)#

#y=e^-x/(x+1)#

The slope #(m)# of the tangent to #y# at #x=1# is #y'(1)#

#y' = (((x+1)*-e^-x) -(e^-x*1))/(x+1)^2#

#= (-e^-x(x+1+1))/(x+1)^2#

#=-(e^-x(x+2))/(x+1)^2#

#:. m=-(e^-1(1+2))/(1+1)^2#

#=(3e^-1)/2^2#

#=-3/(4e)#