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For some reason I can't for the life of me understand total derivative of multivariable function. I understand partial derivatives, you let one variable change and keep the others fixed, but total derivative doesn't make sense. By definition it's the best linear approximation of the function at a given point. So is it a linear transformation? So for example given the function $f(x,y)=x^2+y^2$ I can take the partial derivatives separately and get the function $f(x,y)=2x+2y.$ But if I have understood correctly that is different from the total derivative at point $(x,y).$ Could someone please explain it to me very carefully and simply. Any answers appreciated :)

J. W. Tanner
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Touko Puro
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  • See here https://math.stackexchange.com/questions/3754445/how-can-one-point-determine-a-unique-straight-line-in-differentiation/3754577#3754577. For the gradient, have a look at https://math.stackexchange.com/questions/3159730/dot-product-of-the-gradient-of-a-function/3160777#3160777. – Michael Hoppe Jul 20 '20 at 17:26
  • This paper has a nice review of multivariate differentiation in section $5$. – saulspatz Jul 20 '20 at 17:26
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    The total derivative is a linear transformation. If $f \colon \mathbf R^n \to \mathbf R^m$ is described componentwise as $f(\mathbf x) = (f_1(\mathbf x), \ldots, f_m(\mathbf x))$, for $\mathbf x$ in $\mathbf R^n$, then the total derivative of $f$ at $\mathbf x$ is the $m \times n$ matrix $(\partial f_i/\partial x_j)$ where the partial derivatives are computed at $\mathbf x$. For example, if $f \colon \mathbf R^2 \to \mathbf R$ by $f(x,y) = x^2+y^2$ then the total derivative of $f$ at $(x,y)$ is the $1 \times 2$ matrix $(2x \ \ 2y)$. – KCd Jul 20 '20 at 17:42
  • I’m voting to close this question because OP has in a typical fashion not reacted on comments or answers. – Kurt G. Sep 07 '23 at 09:33

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At least in the special case of $f:\Bbb{R}^n\to \Bbb{R}~;~ f:\mathbf{x}\mapsto f(\mathbf{x})$, the total derivative of $f$ w.r.t an arbitrary variable $u$ is $$\frac{\mathrm{d}f}{\mathrm{d}u}=\sum_{i=1}^n \frac{\partial f}{\partial x_i}\frac{\mathrm{d}x_i}{\mathrm{d}u}$$ This is a straightforward formula, but I can provide some intuition behind it if you want. @KCd 's comment above briefly addresses the more general $\Bbb{R}^n\to\Bbb{R}^m$ case.

K.defaoite
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