The derivative \frac{d{\bf r}}{dt}={\bf r}'(t) of a vector-valued function {\bf r}(t)=(x(t),y(t),z(t)) is defined by \begin{equation}\label{eq:vectorderivative}\frac{d{\bf r}}{dt}=\lim_{\Delta t\to 0}\frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t}\end{equation} In case of a scalar-valued function or a real-valued function, the geometric meaning of derivative is that it is the slope of tangent line. In case of a vector-valued function, the geometric meaning of derivative is that it is the tangent vector. It can be easily seen from Figure 1. As \Delta t gets smaller and smaller, \frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t} gets closer to a line tangent to {\bf r}(t).

From the definition \eqref{eq:vectorderivative}, it is straightforward to show \begin{equation}\label{eq:vectorderivative2}{\bf r}'(t)=(x'(t),y'(t),z'(t))\end{equation}
If {\bf r}'(t)\ne 0, the unit tangent vector {\bf T}(t) of {\bf r}(t) is given by \begin{equation}\label{eq:unittangent}{\bf T}(t)=\frac{{\bf r}'(t)}{|{\bf r}'(t)|}\end{equation}
Example.
- Find the derivative of {\bf r}(t)=(1+t^3){\bf i}+te^{-t}{\bf j}+\sin 2t{\bf k}.
- Find the unit tangent vector when t=0.
Solution.
- Using \eqref{eq:vectorderivative2}, we have {\bf r}'(t)=3t^2{\bf i}+(1-t)e^{-t}{\bf j}+2\cos 2t{\bf k}
- {\bf r}'(0)={\bf j}+2{\bf k} and |{\bf r}'(0)|=\sqrt{5}. So by \eqref{eq:unittangent}, we have {\bf T}(0)=\frac{{\bf r}'(0)}{|{\bf r}'(0)|}=\frac{1}{\sqrt{5}}{\bf j}+\frac{2}{\sqrt{5}}{\bf k}

The following theorem is a summary of differentiation rules for vector-valued functions. We omit the proofs of these rules. They can be proved straightforwardly from differentiation rules for real-valued functions. Note that there are three different types of the product rule or the Leibniz rule for vector-valued functions (rules 3, 4, and 5).
Theorem. Let {\bf u}(t) and {\bf v}(t) be differentiable vector-valued functions, f(t) a scalar function, and c a scalar. Then
- \frac{d}{t}[{\bf u}(t)+{\bf v}(t)]={\bf u}'(t)+{\bf v}'(t)
- \frac{d}{dt}[c{\bf u}(t)]=c{\bf u}'(t)
- \frac{d}{dt}[f(t){\bf u}(t)]=f'(t){\bf u}(t)+f(t){\bf u}'(t)
- \frac{d}{dt}[{\bf u}(t)\cdot{\bf v}(t)]={\bf u}'(t)\cdot{\bf v}(t)+{\bf u}(t)\cdot{\bf v}'(t)
- \frac{d}{dt}[{\bf u}(t)\times{\bf v}(t)]={\bf u}'(t)\times{\bf v}(t)+{\bf u}(t)\times{\bf v}'(t)
- \frac{d}{dt}[{\bf u}(f(t))]=f'(t){\bf u}'(f(t)) (Chain Rule)
Example. Show that if |{\bf r}(t)|=c (a constant), then {\bf r}'(t) is orthogonal to {\bf r}(t) for all t.
Proof. Differentiating {\bf r}(t)\cdot{\bf r}(t)=|{\bf r}(t)|^2=c^2 using the Leibniz rule 5, we obtain 0=\frac{d}{dt}[{\bf r}(t)\cdot{\bf r}(t)]={\bf r}'(t){\bf r}(t)+{\bf r}(t){\bf r}'(t)=2{\bf r}'(t)\cdot{\bf r}(t) This implies that {\bf r}'(t) is orthogonal to {\bf r}(t).
As seen in \eqref{eq:vectorderivative2}, the derivative of a vector-valued functions is obtained by differentiating component-wise. The indefinite and definite integral of a vector-valued function are done similarly by integrating component-wise, namely\begin{equation}\label{eq:vectorintegral}\int{\bf r}(t)dt=\left(\int x(t)dt\right){\bf i}+\left(\int y(t)dt\right){\bf j}+\left(\int z(t)dt\right){\bf k}\end{equation} and \begin{equation}\label{eq:vectorintegral2}\int_a^b{\bf r}(t)dt=\left(\int_a^b x(t)dt\right){\bf i}+\left(\int_a^b y(t)dt\right){\bf j}+\left(\int_a^b z(t)dt\right){\bf k}\end{equation}, respectively. When evaluate the definite integral of a vector-valued function, one can use \eqref{eq:vectorintegral2} but it would be easier to first find the indefinite integral using \eqref{eq:vectorintegral} and then evaluate the definite integral the Fundamental Theorem of Calculus (and yes, the Fundamental Theorem of Calculus still works for vector-valued functions).
Example. Find \int_0^{\frac{\pi}{2}}{\bf r}(t)dt if {\bf r}(t)=2\cos t{\bf i}+\sin t{\bf j}+2t{\bf k}.
Solution. \begin{align*}\int{\bf r}(t)dt&=\left(\int 2\cos tdt\right){\bf i}+\left(\int \sin tdt\right){\bf j}+\left(\int 2tdt\right){\bf k}\\&=2\sin t{\bf i}-\cos t{\bf j}+t^2{\bf k}+{\bf C}\end{align*} where {\bf C} is a vector-valued constant of integration. Now, \int_0^{\frac{\pi}{2}}{\bf r}(t)dt=[2\sin t{\bf i}-\cos t{\bf j}+t^2{\bf k}]_0^{\frac{\pi}{2}}=2{\bf i}+{\bf j}+\frac{\pi^2}{4}{\bf k}
Examples in this note have been taken from [1].
References.
[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole