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

Answer 1

Elijah Abram

#-3/(4e)#

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

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

Applying the quotient rule

#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)#

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Answer 2

Cameron Alexander

To find the slope of a tangent line to the graph of a function at a specific point, you can use the derivative of the function.

To find the derivative of the function y=e^(-x)/(x+1), you can use the quotient rule.

The derivative of y with respect to x is given by:

dy/dx = [(x+1)(-e^(-x)) - e^(-x)(1)] / (x+1)^2

To find the slope of the tangent line at x=1, substitute x=1 into the derivative:

dy/dx = [(1+1)(-e^(-1)) - e^(-1)(1)] / (1+1)^2

Simplifying this expression will give you the slope of the tangent line at x=1.

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Answer from HIX Tutor

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