Fairly simple question, suppose there is a function $f: \mathbb R^n \to \mathbb R$, and a scalar $t \in \mathbb R$.
is it possible to find $D_f(tx)$ using only $t$ and $D_f(x)$? Perhaps using chain rule?
Clarification: $D_f(x)$ is the matrix (or in this case, vector) of partial derivatives at $x$
Can we even say anything about $f(tx)$? if we take it to the realm of $\mathbb R \to \mathbb R$, we can't even know for sure that it is differentiable there.
On
What you call $D_f(x)$ is for all purposes the gradient of $f$ at the point $x$, denoted by $\nabla f(x)$ by many. As a consequence the term $D_f(tx)=\nabla f(tx)$ denotes the gradient of $f$ at the point $tx$. This gradient vector is attached to the point $tx$ far away from $x$. Unless $f$ has special long-range properties the vector $\nabla f(tx)$ is in no way related to the vector $\nabla f(x)$.
It is another story if for given $f:\ {\mathbb R}^n\to{\mathbb R}$ and given scalar $t$ you consider the new function $$g(x):=f(tx)\ .$$ For the derivative of this "contracted copy of $f\,$" the chain rule indeed applies: You can write $$g=f\circ T\ ,$$ where $T$ is given by $T:\ x\mapsto tx$. According to the chain rule we then have $$dg(x)=df(Tx)\circ dT(x)=df(Tx)\circ T=t\ df(Tx)\ ,$$ which can also be written as $$\nabla g(x)=t\ \nabla f(tx)\ .$$
Hint: Let $\varphi : \Bbb{R}\times \Bbb{R}^n \to \Bbb{R}^n$ be defined by $\varphi(t,x) = tx$. Then what you're looking at is the derivative of $f \circ \varphi$. You then have $D(f \circ \varphi)(t,x) = Df(\varphi(t,x)) \circ D\varphi(t,x)$ by the chain rule.