If a function
\[
f(x)=
\begin{cases}
\dfrac{a}{|x|}, & x\le -1 \text{ or } x\ge 1,[6pt] \\
x^2+b, & -1<x<1,
\end{cases}
\]
is differentiable on \(\mathbb{R}\), then \(a+b=\)
Show Hint
Whenever a piecewise function is said to be differentiable, first apply continuity and then equate the left-hand and right-hand derivatives. Missing the continuity condition is the most common mistake.