By definition of absolute value, you have

or more simply,

On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,

Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:


All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.
It is different because you are doing a different method to complete the problem. It is the same because it ends up with the same answer, provided you complete the process correctly.
Let
x---------> amount of money saved by Mark
x---------> amount of money saved by Neil
we know that
x=y+258-------> equation 1
x/y=(7/4)-------> equation 2
substitute equation 1 in equation 2
(y+258) /y=7/4--------> 4*(y+258)=7*y------> 4y+1032=7y
7y-4y=1032--------> 3y=1032------> y=$344
x=344+258------>x=$602
the answer is
$602