Question 976797

{{{(d/dx)(abs(x-1)+abs(x-3)) = (x-3)/(abs(x-3))+(x-1)/(abs(x-1))}}}

if {{{x=2}}}, we have

{{{(d/dx)(abs(x-1)+abs(x-3)) = (2-3)/(abs(2-3))+(2-1)/(abs(2-1))}}}

{{{(d/dx)(abs(x-1)+abs(x-3)) = -1/(abs(-1))+1/(abs(1))}}}


{{{(d/dx)(abs(x-1)+abs(x-3)) = -1/1+1/1}}}

{{{(d/dx)(abs(x-1)+abs(x-3)) = -1+1}}}

{{{(d/dx)(abs(x-1)+abs(x-3)) = 0}}}

so, your answer is b){{{0}}}