Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - Review Exercises - Page 393: 13

$$y = -x+1$$

Let the equation of the tangent be $$y=mx+c$$ We know that the slope $m =\frac{dy}{dx}|_{x=-1}$ at x=-1 Or $m=\frac{1}{2+x}-\frac{2}{(2+x)^2}|_{x=-1} = -1$ Therefore, $y=-x+C$ But the tangent must pass through the given point. So, (-1, 2) is a point on the tangent. Therefore, $2 = -(-1)+C$ or $C=1$. Thus $$y = -x+1$$ is the tangent of the given function at the given point/