Let $f(x) =|x|+|x+1|+...+|x+99|$ then find: $f'\left(\frac{-9}{2}\right)=?$ 
Let $f(x) =|x|+|x+1|+...+|x+99|$ then find :
$$f'\left(\frac{-9}{2}\right)=?$$


My Try :
$$\lim_{x\to \frac{-9}{2}}\dfrac{f(x)-f\left(\frac{-9}{2}\right)}{\left(x+\frac{9}{2}\right)}
=
\lim_{x\to \frac{-9}{2}}\dfrac{(90x+95(52)-10)-f\left(\frac{-9}{2}\right)}
{\left(x+\frac{9}{2}\right)}
=?
$$
Now what ?

we have $f(x)=\left| x\right| +\left| x+1\right| +\left| x+2\right| +\left| x+3\right| +\left| x+4\right|+\ldots +\left| x+99\right|$
In the interval $\left[-4.6,-4.4\right]$ the function is
$f(x)=-x-x-1-x-2-x-3-x-4+(x+5)+(x+6)+(x+7)+(x+8)+(x+9) +\ldots+x+99$
the first $10$ terms containing $-x$ and $x$ simplify to $25$ while the other $90$ give
$f(x)=90x+M$
differentiating will give $f'(-9/2)=90$
Hope this helps
