---- Let $\Omega \subset \mathbb{R} ^{m}$ be [[open set|open in]] $\mathbb{R}^{m}$. Let $f:\Omega \to \mathbb{R} ^{n}$ be [[derivative|differentiable]] at $p \in \Omega$. > [!theorem] Theorem. ([[total derivative equals Jacobian]]) > Then the [[derivative]] of $f$ at $\vec p$, $Df(\vec p)$, equals the [[Jacobian]] $J$ of $f$ at $\vec p$: $D\vec f(\vec p)=J = \begin{bmatrix} \frac{ \partial f_{1} }{ \partial x_{1} } & \dots & \frac{ \partial f_{1} }{ \partial x_{m} } \\ \vdots & \ddots & \vdots \\ \frac{ \partial f_{n} }{ \partial x_{1} } & \dots & \frac{ \partial f_{n} }{ \partial x_{m} } \end{bmatrix} = \begin{bmatrix} D_{1}f_{1}(\vec p) & \dots & D_{m}f_{1}(\vec p) \\ \vdots & \ddots & \vdots \\ D_{1}f_{n}(\vec p) & \dots & D_{m}f_{n}(\vec p) \end{bmatrix}.$ > [!proof]- Proof. ([[total derivative equals Jacobian]]) > *Proof*: Suppose $\vec f$ is differentiable at $\vec p$ and with this write $B=D\vec f(\vec p)$. By the **preliminary lemma**, we have the [[directional derivative]] of $\vec f$ at $\vec e_{j}$ (the $j^{th}$ [[partial derivative]] of $\vec f$) is differentiable at $\vec p$ and $f'(\vec p; \vec e_{j}) = D\vec f(\vec p) \vec e_{j} = B\vec e_{j}. $ But think about this— when you multiply a [[matrix]] by a standard [[basis]] [[vector]], the result is 'where that basis vector lands after the [[linear map]]', which will of course end up yielding as the product the corresponding column of the matrix. But *where that basis vector ends up after the linear transformation*... well, we've just shown that in the case of $\vec e_{j}$ this is $f'(\vec p; \vec e_{j})$ (*see [[existence of derivative guarantees existence of all directional derivatives]]s with $f'(\vec a; \vec u) = Df(\vec a)\cdot \vec u$*). So we can write $B = \begin{bmatrix}\vec f'(\vec p;\vec e_{1}) & \dots & \vec f'(\vec p;\vec e_{m})\end{bmatrix}.$ We can then expand $f'(\vec p; \vec e_{j})$ into its components, since it is (or can be) [[vector]]-valued. This yields $B=\begin{bmatrix} \frac{ \partial f_{1} }{ \partial x_{1} } & \dots & \frac{ \partial f_{1} }{ \partial x_{m} } \\ \vdots & \ddots & \vdots \\ \frac{ \partial f_{n} }{ \partial x_{1} } & \dots & \frac{ \partial f_{n} }{ \partial x_{m} } \end{bmatrix} = \begin{bmatrix} D_{1}f_{1}(\vec p) & \dots & D_{m}f_{1}(\vec p) \\ \vdots & \ddots & \vdots \\ D_{1}f_{n}(\vec p) & \dots & D_{m}f_{n}(\vec p) \end{bmatrix},$ which is what we wanted to show! ![[CleanShot 2022-09-11 at 10.24.14.jpg]] #### References ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```