How do I go about calculating the derivative of this curve from $\mathbb{R}$ to $\mathbb{R}^3$?
$$γ(t) = (t\cdot\cos(2t), t\cdot\sin(2t), t)$$
I have tried simply taking the derivative of the three components, and adding them, but I'm not sure if this is the correct way of doing it.
On
As mentioned by Hyperion, for a multidimensional equation of this form there are several different types of derivative. We can visualise gamma as $\gamma=\gamma(x(t),y(t),z(t))$.
You may be referring to the total derivative, which has the definition: $$\frac{d\gamma}{dt}=\frac{\partial\gamma}{\partial x}\frac{dx}{dt}+\frac{\partial\gamma}{\partial y}\frac{dy}{dt}+\frac{\partial\gamma}{\partial z}\frac{dz}{dt}$$ since we have the function defined: $$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}t\cos(2t)\\t\sin(2t)\\t\end{pmatrix}$$ we are able to show that: $$\frac{d}{dt}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\cos(2t)-2t\sin(2t)\\\sin(2t)+2t\cos(2t)\\1\end{pmatrix}$$ and this can then be substituted into the formula.
You did one step too much. Just differentiate each component and stop there. The derivative ought to be a new function $\Bbb R\to\Bbb R^3$.
Intuitively, the derivative measures how much the value of $\gamma$ changes. The values of $\gamma$ are in $\Bbb R^3$, so changes in the value of $\gamma$ are also in $\Bbb R^3$.
More rigorously, given a function $\gamma:\Bbb R\to\Bbb R^3$, with component functions $\gamma_1,\gamma_2,\gamma_3$, its derivative is by definition $$ \gamma'(t)=\lim_{h\to\infty}\frac{\gamma(t+h)-\gamma(t)}h\\ =\lim_{h\to0}\left(\frac{\gamma_1(t+h)-\gamma_1}h, \frac{\gamma_2(t+h)-\gamma_2}h, \frac{\gamma_3(t+h)-\gamma_3}h\right)\\ =(\gamma'_1(t),\gamma_2'(t),\gamma'_3(t)) $$ If you want the speed of $\gamma$, rather than the velocity, then that's simply the length of this vector. You find this the way you usually find the length of a vector: not by adding the components, but by the Pythagorean theorem.