If the derivative of the function $f(x) = \begin{cases} ax^2 + b & \text{if } x < -1 \\ bx^2 + ax + 4 & \text{if } x \geq -1 \end{cases}$ is continuous everywhere then

For $f(x)$ to have a continuous derivative everywhere, $f(x)$ must be differentiable and hence continuous at $x = -1$.

Equating the left-hand and right-hand limits of $f(x)$ at $x = -1$:

$a(-1)^2 + b = b(-1)^2 + a(-1) + 4$

$a + b = b - a + 4$

$2a = 4 \Rightarrow a = 2$

The derivative of the function is:

$f'(x) = \begin{cases} 2ax & \text{if } x < -1 \\ 2bx + a & \text{if } x > -1 \end{cases}$

Since $f'(x)$ is continuous at $x = -1$, its left-hand and right-hand limits at $x = -1$ must be equal:

$2a(-1) = 2b(-1) + a$

$-2a = -2b + a$

$2b = 3a$

Substituting $a = 2$:

$2b = 3(2) \Rightarrow b = 3$

Answer: $a = 2, b = 3$