Let \(R\) be a ring, \(S\) a nonempty subset of \(R\), and \(Z\) the center of \(R\). For \(x,y\in R\) denote \(xy- yx\) by \([x, y]\) and \(xy + yx\) by \(x\circ y\). Let \(d\) be a derivation on \(R\). For prime \(R\) and \(S\) either an ideal or a Lie ideal, the authors study commutativity under the assumption that one of the following holds for all \(x,y\in S\): (i) \(d([x, y])= [x, y]\), (ii) \(d(x\circ y)= x\circ y\), (iii) \(d(x)\circ d(y)= 0\), (iv) \(d(x)\circ d(y)= x\circ y\). The following results are typical: (A) Let \(R\) be a prime ring with \(\operatorname{char} (R)\neq 2\) and \(U\) a nonzero Lie ideal. If \(R\) admits a derivation \(d\) such that \(d([u, v])= [u, v]\) for all \(u,v\in U\), then \(U\subseteq Z\); (B) Let \(R\) be a prime ring and \(I\) a nonzero ideal of \(R\). If \(R\) admits a derivation \(d\) such that \(d(x\circ y)= x\circ y\) for all \(x, y\in I\), then \(R\) is commutative.