- #1
shaggymoods
- 26
- 0
Hey guys, so this may be a really silly question, but I'm trying to grasp a subtle point about higher-order derivatives of multivariable functions. In particular, suppose we have an infinitely differentiable function
[tex]f: \mathbb{R}^{n} \rightarrow \mathbb{R}[/tex]
I know that the first derivative of this function is a linear map [tex]\lambda: \mathbb{R}^{n}\rightarrow\mathbb{R}[/tex]. However, when we take the second-derivative of [tex]\lambda[/tex], some questions arise for me:
1.) If we are taking this derivative when considering [tex]\lambda[/tex] as a linear function, then we'd just get back [tex]\lambda[/tex], which isn't the case. So how are we interpreting the first derivative when taking a second?
2.) In general, why do we say that [tex]D^{k}f:\mathbb{R}^{n^{k}}\rightarrow\mathbb{R}[/tex] and not [tex]D^{k}f:\mathbb{R}^{n}\rightarrow\mathbb{R}[/tex] ??
Thanks in advance.
[tex]f: \mathbb{R}^{n} \rightarrow \mathbb{R}[/tex]
I know that the first derivative of this function is a linear map [tex]\lambda: \mathbb{R}^{n}\rightarrow\mathbb{R}[/tex]. However, when we take the second-derivative of [tex]\lambda[/tex], some questions arise for me:
1.) If we are taking this derivative when considering [tex]\lambda[/tex] as a linear function, then we'd just get back [tex]\lambda[/tex], which isn't the case. So how are we interpreting the first derivative when taking a second?
2.) In general, why do we say that [tex]D^{k}f:\mathbb{R}^{n^{k}}\rightarrow\mathbb{R}[/tex] and not [tex]D^{k}f:\mathbb{R}^{n}\rightarrow\mathbb{R}[/tex] ??
Thanks in advance.