![\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}](/tpl/images/0381/2308/8ec0b.png)
Did i do this right?
find the derivative of f(x) = |x + 2| at the point (1, 3).
...
Mathematics, 19.11.2019 12:31, paige3673
Did i do this right?
find the derivative of f(x) = |x + 2| at the point (1, 3).
a. -1
b. 0
c. 1
d. does not exist.
i found the derivative which is (x+2)/|x + 2|. then, i plugged 1 in for x and got 3/3, which is just 1. the answer is c.
Answers: 1
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