How do you get the derivative of absolute values ? Im trying to derive f(x)=3x^3+16|x-1| so that i can find the points where the derivative is 0

Question

amistre64 · Answer

you can split the absolute value up into 2 parts; and remember that the tip of the graph has no derivative.

d/dx (|x|) = d/dx (x), for x >0  or
                  d/dx (-x), for x < 0
                   d/dx, DNE at x=0

amistre64 · Answer

another common, and prolly more useful method is explained in this link: http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/calculus/DerivativesInvolvingAbsoluteValue.pdf

anonymous · Answer

attachment

anonymous · Answer

So the derivative of the original function is:

$$9x^2-16(x-1)(1)/\sqrt{(x-1)^2}$$

anonymous · Answer

If you set this equal to 0 and solve, you end up with +/4/3 and only the negative answer is meaningful. (see graph of function in attachment above.) The last step is to look at the discontinuous point  (x=1) and decide if is a local min or max.