Question

Did I do this right?

Answer 1

$\bf f(x)=|x+2|\implies f(x)=\pm\sqrt{(x+2)^2}\implies f(x)=\left[ (x+2)^2 \right]^{\frac{1}{2}} \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{2}[(x+2)^2]^{-\frac{1}{2}}\cdot 2(x+2)^1\cdot 1}\implies \cfrac{dy}{dx}=[(x+2)^2]^{-\frac{1}{2}}(x+2) \\\\\\ \cfrac{dy}{dx}=\cfrac{x+2}{[(x+2)^2]^{\frac{1}{2}}}\implies \left. \cfrac{dy}{dx}=\cfrac{x+2}{\pm \sqrt{(x+2)^2}} \right|_{x=1}\implies \cfrac{3}{\pm 3}$

check the picture below. the point 1,3 is on the increasing slope side, so will be 3/3 or 1.