Yes! The density of a continuous distribution is the derivative of the CDF.$^{\dagger}$
Example: the uniform distribution, say on $(0,1)$, which has PDF $f(x)= \left\{
\begin{array}{ll}
1 & x\in (0,1)\\
0 & x\notin (0,1) \\
\end{array}
\right.$.
Then the CDF is $F(x)= \left\{
\begin{array}{ll}
0& x\le0\\
x & x\in (0,1)\\
1 & x\ge0) \\
\end{array}
\right.$.
You can see that $\dfrac{dF(x)}{dx} = f(x)$, as you'd expect.
We don't usually talk about the PDF as being continuous, however. Continuous vs discrete concerns the CDF. Fair warning: the details of this quickly get you into heavy real analysis, including measure theory.
$^{\dagger}$In some sense, you always can get the density through a derivative. Measure theory unifies discrete, continuous, and even funkier distributions and gives their densities through something called the Radon-Nikodym derivative.