Question

what is the slope of the line tangent to the graph of $y = \frac{e^{-x}}{x + 1}$ at $x = 1$?
a $-\frac{1}{2}$
b $-\frac{3}{4e}$
c $-\frac{1}{e}$
d $\frac{1}{2}$

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if \(y = \frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here, \(u = e^{-x}\), \(u^\prime=-e^{-x}\), \(v=x + 1\), \(v^\prime = 1\). So \(y^\prime=\frac{-e^{-x}(x + 1)-e^{-x}\times1}{(x + 1)^{2}}=\frac{-e^{-x}(x+2)}{(x + 1)^{2}}\).

Step2: Substitute \(x = 1\)

Substitute \(x = 1\) into \(y^\prime\). We get \(y^\prime|_{x = 1}=\frac{-e^{-1}(1 + 2)}{(1+1)^{2}}=\frac{-3e^{-1}}{4}=-\frac{3}{4e}\).

Answer:

B. \(-\frac{3}{4e}\)