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Take for example $f(x,y) = x^y$. I defined the total derivative to be the best linear approximation of $f$. Without working out the Jacobian I found that $$Df(x,y)(h_1,h_2) = h_1yx^{y-1} + h_2x^y\log(x)$$

However the Jacobian gives me a $1 \times 2$ matris: $$\begin{bmatrix} yx^{y-1} & x^y \log (x) \end{bmatrix} $$

I don't really understand how computing the total derivative explicitly gives me a real number, and computing the Jacobian gives me a matrix, how are they related because I do know they somehow are.

helppls
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  • the total derivative of $f$ in $(x,y)$ (which is a linear map!) applied to $(h_1,h_2)$ can be written as matrix multiplication by the means of the jacobian. (Multiply your Jacobian with $[h_1,h_2]^T$ and compare with your first result.) hope that helps you understand the difference between both objects. – Max Oct 20 '15 at 23:26

2 Answers2

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The Jacobian is just a linear function. Apply it to the point $\begin{bmatrix} h_1 \\ h_2\end{bmatrix}$ and you get back the total derivative.

user282321
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    And what are matrices except the representations of linear functions? – user282321 Oct 20 '15 at 23:27
  • matrices give a nice possibility to represent linear maps by matrix multiplication to represent linear functions. but that implies you chose a base. (not all finite dim real vector spaces are $\mathbb{R}^n$, but they are isomorphic to one.) a matrix is a matrix and not a function. you dont apply it to something, you matrix-multiply it with something. – Max Oct 20 '15 at 23:30
  • A basis has been chosen, that's why changing the coordinates changes the Jacobian matrix. And it's common language to say "apply a matrix to another matrix" to mean matrix multiply them. – user282321 Oct 20 '15 at 23:38
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The total derivative is the best linear approximation of $f$, thus an affine plane in $x,y$ space.

$$ f(x+h_1,y+h_2) = f(x, y) + \mbox{grad } f \cdot (h_1, h_2)^T + O(h^2) $$

Compare this with the $f' : \mathbb{R}^2 \to \mathbb{R}$ in $$ \lim_{(h_1, h_2) \to 0} \frac{\lVert f(x+h_1, y+h_2) - f(x,y) - f'(x,y) (h_1, h_2)^T\rVert}{\lVert (h_1, h_2)^T \rVert} = 0 $$ You computed $$ \Delta f \approx \mbox{grad } f \cdot (h_1, h_2)^T $$ while the total derivative is $$ f' = (\mbox{grad } f)^T $$ in this case.

mvw
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  • how can the total derivative be a matrix, it is a scalar in this case as $f:R^2-> R$ so does $D$ – helppls Oct 21 '15 at 00:19
  • The total derivative concept exists in two flavours, as $df/dt = \mbox{grad } f \cdot dr/dt$ which is a scalar valued function, or as linear map, here a linear form, the transpose of the gradient. – mvw Oct 21 '15 at 00:31