I'm reading a paper about translating the traditional notation for multivariable derivatives into tensor products.
Let $f,g$ be functions from $U\subseteq\mathbb{R}^n$ that map to a finite dimensional space $V$. $D(fg)=Df(I\otimes g)+fDg$, where $I$ is the identity map on $V$. This expression can be differentiated again, using $D(I\otimes g)=(DI\otimes g)+(I\otimes g)=(0\otimes g)+(I\otimes g)=I\otimes g$
Why is $DI=0?$ Shouldn't it be $I$, since the derivative of a linear operator is itself?