I have a function $\phi: \mathbb{R} \rightarrow \mathbb{R}$ that is a test function and $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is defined by $f(x) = \phi(\frac{\|x\|}{n})$.
Now if I take any partial derivative with respect to any coordinate $x_j$ I get $$|\partial_j f(x) | \le \left|\partial_j \phi\left(\frac{\|x\|}{n}\right)\right| \le \|\phi'\|_\infty \frac{1}{n}.$$
Now let $D^\alpha$ denote a multiindex derivative (so higher orders are possible and differentiation with respect to mixed coordinates), is there also a way to give such a nice bound for $|D^{\alpha} f(x)|$ in terms of sup-norms of $\phi$ derivatives? Thus, I want to generalize $|\partial_j f(x) | \le |\partial_j \phi\left(\frac{\|x\|}{n}\right)| \le \|\phi'\|_\infty \frac{1}{n}$.
If anything is unclear, please let me know.
This is a quick observation I am not sure this is what you want.
If you go on and compute second derivative, you will have $$\partial_j\partial_i f(x)=\phi''(|x|/n)\frac{1}{n^2}\frac{x_ix_j}{|x|^2} +\phi'(|x|/n)\frac{\delta_{ij}|x|-x_ix_j}{|x|^3}\frac{1}{n} $$
Hence, by induction, you could have $$ |D^\alpha f(x)|\leq \sum_{i=1}^k \|\phi^i\|_{L^\infty}\frac{1}{n^i} $$ where $|\alpha|=k$