Relationship between scalar function by vector differentiation and vector by scalar function differentiation

Question

The following question is regarding matrix calculus. Let $f$ be a function denoted by, $$f = \sqrt{(a_1-a_2)^2+(a_2-a_3)^2+(a_1-a_3)^2+3(a_4^2+a_5^2+a_6^2)}$$ and $\pmb A$ be a vector defined by, $$ \pmb A = (a_1 \space a_2 \space a_3 \space ...a_6)^T$$ $\frac{d f}{d\pmb A}$ can be readily computed by evaluating the following, $$\frac{d f}{d\pmb A} = (\frac{df}{da_1} \space \frac{df}{da_2} \space \frac{df}{da_3} \space ...\frac{df}{da_6})$$ But what I require is an expression for / need to evaluate is $\frac{d\pmb A}{df}$. If $\frac{d\pmb A}{df}$ is, $$\frac{d\pmb A}{df} = (\frac{da_1}{df} \space \frac{da_2}{df} \space \frac{da_3}{df} \space ...\frac{da_6}{df})^T$$ does the following relationship (Inverse function theorem like relationship) hold true for $\frac{d\pmb A}{df}$? $$\frac{d\pmb A}{df} = (\frac{1}{{df} \over {da_1}} \space \frac{1}{{df} \over {da_2}} \space \frac{1}{{df} \over {da_3}} \space ...\frac{1}{{df} \over {da_6}})^T$$ (Here, $\frac{df}{da_i} \neq 0 $ for $i = 1, 2, ..., 6$ and $f$ is continuously differentiable in the interval of interest).

Answer 1

It seems that your assumption should hold. When differentiating a matrix by a scalar quantity, we don't need to worry about how differentiation will affect the dimensionality of our result. In general, given a matrix $X$ we have that

$$ \frac{d}{dt} \left [ \begin{array}{cccc} x_{11} & x_{12} & \ldots & x_{1n} \\ x_{21} & x_{22} & \ldots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} & x_{m2} & \ldots & x_{mn} \\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{cccc} \frac{dx_{11}}{dt} & \frac{dx_{12}}{dt} & \ldots &\frac{dx_{1n}}{dt} \\ \frac{dx_{21}}{dt} & \frac{dx_{22}}{dt} & \ldots &\frac{dx_{2n}}{dt} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{dx_{m1}}{dt} & \frac{dx_{m2}}{dt} & \ldots &\frac{dx_{mn}}{dt} \\ \end{array} \right ]. $$

In the case you have in your prompt, you would simply be taking derivatives $\frac{\partial a_i}{\partial f}$, which can be found implicitly using single variable calculus.