# Coordinate Derivative of a Tensor Product

I know similar questions have been asked on the stack exchange, but all the ones I've found haven't had answers.

I have just started working with tensor products, but have had to learn them a little on the run. I was wondering how the derivative acts on a tensor product.

That is, what is: $$\frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big)$$ ?

Is it: $$\frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big) = \Big( \frac{\mathrm{d}}{\mathrm{d}x} f \left( x \right) \Big) \otimes g \left( x \right) + f \left( x \right) \otimes \Big( \frac{\mathrm{d}}{\mathrm{d}x} g \left( x \right) \Big)$$ or $$\frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big) = \Big( \frac{\mathrm{d}}{\mathrm{d}x} f \left( x \right) \Big) \otimes \Big( \frac{\mathrm{d}}{\mathrm{d}x} g \left( x \right) \Big)$$ or something else?

• Are you sure this is well-defined? Are $f(x)$ and $g(x)$ differentiable functions on $\mathbb R^n$ and $\mathbb R^m$? – Aniruddh Agarwal May 2 '19 at 0:25
• So, tensor product is a product, so the product rule must apply. But let's be a bit more specific. Where are things mapping from and to here? – Ted Shifrin May 2 '19 at 0:34
• @Ted Shifrin I'm not quite sure what detail it would be useful to add to my question here. In actual fact the situation in which I came across this question was dealing with the time derivative of the tensor product of quantum states: $\partial_t \left( \vert \varphi \rangle \otimes \vert \phi \rangle \right)$ so I'm working in Fock space. However this added detail seemed to just confuse the point I was wanting to clarify. – leob May 2 '19 at 3:37
• Right. It's still a product, so the product rule applies. – Ted Shifrin May 2 '19 at 4:29
• Crossposted to physics.stackexchange.com/q/477308/2451 – Qmechanic May 2 '19 at 10:18