# What is the slope of the tangent line to the graph of #y=e^-x/x+1# at x=1?

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To find the slope of the tangent line to the graph of (y = \frac{e^{-x}}{x + 1}) at (x = 1), we first find the derivative of the function, (y), with respect to (x), and then evaluate it at (x = 1).

The derivative of the function (y) with respect to (x) is given by:

[y' = \frac{d}{dx}\left(\frac{e^{-x}}{x + 1}\right)]

Using the quotient rule, the derivative is:

[y' = \frac{(x+1)(-e^{-x}) - e^{-x}(1)}{(x+1)^2}]

Now, substitute (x = 1) into this expression to find the slope of the tangent line at (x = 1).

[y'(1) = \frac{(1+1)(-e^{-1}) - e^{-1}(1)}{(1+1)^2}]

[y'(1) = \frac{-2e^{-1} - e^{-1}}{4}]

[y'(1) = \frac{-3e^{-1}}{4}]

Therefore, the slope of the tangent line to the graph of (y = \frac{e^{-x}}{x + 1}) at (x = 1) is (-\frac{3}{4}e^{-1}).

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