what is the slope of the line tangent to the graph of $y = \\frac{e^{-x}}{x + 1}$ at $x = 1$?\na…

what is the slope of the line tangent to the graph of $y = \\frac{e^{-x}}{x + 1}$ at $x = 1$?\na $-\\frac{1}{2}$\nb $-\\frac{3}{4e}$\nc $-\\frac{1}{e}$\nd $\\frac{1}{2}$

what is the slope of the line tangent to the graph of $y = \\frac{e^{-x}}{x + 1}$ at $x = 1$?\na $-\\frac{1}{2}$\nb $-\\frac{3}{4e}$\nc $-\\frac{1}{e}$\nd $\\frac{1}{2}$

Answer

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})