I've a triangle ABC. Where AC is the hypotenuse and the angle ABC is 90 degress.

AB is $15 km$ and changes with a speed of $600 km/h$.
BC is $5 km$ and changes with a speed of $0 km/h$.
At what speed changes the angle CAB in terms of $rad/h?$

I call AB for $x(t)$ and BC for $y(t)$. Then I know:

$$x(t) = 15 km$$
$$x'(t) = 600 km/h$$
$$y(t) = 5 km$$
$$y'(t) = 0 km/h$$

To find the angle CAB (now called $\theta$):

$$ tan(\theta) = \frac{y(t)}{x(t)} \leftrightarrow \theta = arctan\left(\frac{y(t)}{x(t)}\right)$$.

I derivate this to get the change of $\theta$ in $rad/h$.

$$ \frac{1}{1+\left(\large\frac{y(t)}{x(t)}\right)^2} \times \left(\frac{y'(t)x(t)-y(t)x'(t)}{x(t)^2}\right) = 0.0075~rad/h$$

I know that the answer should be $12~rad/h$ so somewhere I'm wrong. But where?
![enter image description here](https://i.stack.imgur.com/e699q.jpg)