Question

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

Answer 1

The derivative is `y'=\frac{x\cdot e^{-x}\cdot (-1)-e^{-x}\cdot 1}{x^{2}}=\frac{-e^{-x}(x+1)}{x^{2}}`

Therefore the slope of the tangent line at `x=1` is `y'(1)=\frac{-2e^{-1}}{1^{2}}=-2/e\approx -0.7358`.

For extra fun, since `y(1)=\frac{e^{-1}}{1}+1=\frac{1+e}{e}`, the equation of the tangent line at `x=1` is `y=\frac{1+e}{e}-\frac{2}{e}(x-1)=-\frac{2}{e}x+\frac{3+e}{e}\approx -0.7358x+2.1036`.