# product rule for matrix functions?

Given a real rectangular matrix $X$, and two scalar-valued matrix functions, $f(X)$ and $g(X)$, does the product rule for differentiation of a product of scalar valued functions, hold when differentiating the product, $f(X)g(X)$ w.r.t $X$ ? If not, what would be the corresponding product rule? Let's assume that the product $f(X)g(X)$ gives a real valued scalar, and is well-defined in terms of the dimensions.

Note: $f(.)$ and $g(.)$ can be the matrix trace function for example.

Yes, the standard product rule applies. The gradient of the product is $$f(X)\nabla_X g(X)+g(X)\nabla_X f(X).$$ The dimensions of the gradients, of course, are the same as those of $X$ itself.
The product rule holds in very great generality. Let $$X,Y,Z,W$$ be Banach spaces with open subset $$U \subset X$$, and suppose $$f: U \rightarrow Y$$ and $$g: U \rightarrow Z$$ are Frechet differentiable. If $$B(\cdot, \cdot): Y \times Z \rightarrow W$$ is a continuous bilinear map, then for any $$\xi \in X$$,
$$\frac{d}{dx}[ B(f(x), g(x))](\xi) = B(f'(x)\xi, g(x)) + B(f(x), g'(x)\xi)$$
where all the derivatives in question are Frechet derivatives. To apply to your case, we take $$U = X = \mathbb{R}^{n \times n}$$, $$Y =Z = W = \mathbb{R}$$, and $$B(y,z) = yz$$.
• This notation is a little imprecise, since we write $B(f'(x), g(x))$ where $f'(x)$ is not an element of $Y$ but rather of $L(X,Y)$. So I guess we are meant to understand $B(f'(x), g(x))$ as the linear operator $\xi \mapsto B(f'(x) \xi, g(x))$ in $L(X,W)$. – Nate Eldredge Apr 25 '20 at 15:06