Show that the derivative $\theta(t)$ is continuous I am making a problem, and I need to show that the derivative of the function $\theta(t)$ is continuous. $\theta(t)$ is defined as follows:

Let $f$ be a quadratic function, and $\theta(t)$ is the angle between the line segment between $(0, 0)$ and $(t, f(t))$, and the positive x-axis.

This is a part of my solution.

$\tan\theta(t)=\frac{f(t)}{t}$ for all $t \neq 0$, and by differentiating each side, we get
$$\theta'(t)\times\sec^2\theta(t)=\frac{tf'(t)-f(t)}{t^2}$$
As $\sec^2\theta(t)=1+\tan^2\theta(t)$, by plugging in the original equation, we get
$$\theta'(t)=\frac{tf'(t)-f(t)}{(f(t))^2+t^2}$$

However, this method only works for $t\neq 0$, since if $t=0$, then $\theta(0)=\frac{\pi}{2}$ so  $\tan \theta(0)$ is not defined. I need to show that this function has a continuous derivative, in order to make the problem to not have an unexpected error. I suspect that by plugging in $t=0$ to $\theta'(t)$ will give us the value for $\theta'(0)$, but I have no mathematical reason for that. Can I get any help?

The complete definition of $\theta (t)$ should be the following:
$$\begin{aligned}\theta(t)&=\tan^{-1}\left(\frac {f(t)}{t}\right), tf(t)>0\\&=\pi+\tan^{-1}\left(\frac {f(t)}{t}\right), tf(t)<0\\&=\frac {\pi}{2}, t=0\end{aligned}$$
It can be easily verified that $\theta(t)$ is differentiable throughout, since $\theta(t)$ is continous at $t=0$ and:
$$\lim_{t\to 0^{-}}\theta'(t)=\lim_{t\to 0^{+}} \theta'(t)=-\frac {1}{f(0)}$$
Since $\theta(t)$ is differentiable at $t=0$, this means that $$\theta'(0)=\lim_{t\to 0} \theta'(t)=-\frac {1}{f(0)}$$
This means that $\theta'(t)$ is continous at $t=0$.
