Do nonsingular points get mapped to nonsingular points in a branched cover

Let $\pi:C\to D$ be a finite surjective morphism of noetherian integral schemes. Let $x\in C$ be a nonsingular point. Does it follow that $\pi(x)$ is nonsingular?

What if we impose some conditions like normality or regularity?

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(Let me add that you only need flatness to make the conclusion. That is, a flat morphism of noetherian schemes $X \to Y$ sends nonsingular points to nonsingular points. This is a consequence on Serre's characterization of regular local rings as those with finite global dimension. So if $A \to B$ is a local, flat, homomorphism of local noetherian rings, then if $B$ is regular, so is $A$. To see this, let $M$ be any $A$-module, and choose an infinite free resolution $P_\bullet$ of $M$ by finitely generated modules. After tensoring with $B$, we get an infinite free resolution of $M \otimes_A B$, i.e. $P_\bullet \otimes_A B$. The kernel at some point of $P_\bullet \otimes_A B$ will be free because $B$ has finite global dimension. That means the kernel at some point of $P_\bullet$ will be flat, hence free. So $A$ has finite global dimension.)