Partials of $f(t+\theta)$ evaluated at $t=0$ versus partials of $f(\theta)$

Question

Suppose that $f:\mathbb{R}^n\to\mathbb{R}$ is infinitely differentiable with respect to its components. For $k\geq 1$ and $1\leq i_1,\ldots,i_k\leq n$, let's look at a $k$-th order mixed partial of a particular form: $$ A\equiv\frac{\partial^k f(t+\theta)}{\partial t_{i_1}\ldots \partial t_{i_k}}\Bigg|_{t=0}. $$ I'm trying to justify properly that the above expression equals $$ B\equiv\frac{\partial^k f(\theta)}{\partial \theta_{i_1}\ldots \partial \theta_{i_k}}. $$ But I'm being confused by the multivariate nature of the domain of $f$. Indeed, when $n=1$, the problem is simple $$ A=f^{(k)}(t+\theta)\Big|_{t=0}=f^{(k)}(\theta)=B. $$ Can you please help me? Edit: I think I got it. Can you please verify my answer below (posted as wiki)?

Answer 1

Fix $k\geq 1$ and $1\leq i_1,\ldots,i_k\leq n$. Define the function $g:\mathbb{R}^n\to\mathbb{R}$ as follows $$ g(z)=\frac{\partial^kf(z)}{\partial z_{i_1}\ldots\partial z_{i_k}}. $$ Then it's immediate that $B=g(\theta)$. Let's look at $A$. By the Chain Rule, $$ \frac{\partial f(t+\theta)}{\partial t_{i_k}}=\frac{\partial f(t+\theta)}{\partial(t_{i_k}+\theta_{i_k})}\frac{\partial(t_{i_k}+\theta_{i_k})}{\partial t_{i_k}}=\frac{\partial f(t+\theta)}{\partial(t_{i_k}+\theta_{i_k})}. $$ Proceeding like this, we get $$ \frac{\partial^k f(t+\theta)}{\partial t_{i_1}\ldots \partial t_{i_k}}=\frac{\partial^k f(t+\theta)}{\partial (t+\theta)_{i_1}\ldots \partial (t+\theta)_{i_k}}=g(t+\theta). $$ Thus, the original claim that $A=B$ is equivalent to $$ g(t+\theta)\Big|_{t=0}=g(\theta) $$ which holds trivially.