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Let $f(x_1, ..., x_n)$ be a function defined in a some area of the point $x^0 \in \mathbf R^n$ and let the functions of one variable $\phi_1(t), ..., \phi_n(t)$ are defied and differentiable around some point $t^0 \in \mathbf R$. We also know that $x_i^0=\phi_i(t^0)$ for $1 \le i \le n$.

Let's assume that $f(x)$ is differiable in $x^0$, and $\phi_i(t)$ in $t^0$.

Then the composite function $F(t)=f(\phi_1(t),...,\phi_n(t))$ is differentiable and

$$ F'(t)=\sum_{i=1}^{n}{\frac{\partial f}{\partial x_i}(x^0)\phi_i'(t^0)} $$

OK, so what is the intuition behind this? Why are we summing?

My understanding is that this is a dot product of the vector of he derivatives of the outer functions with the vector of the derivatives of the inner functions $(\phi_1(t), ..., \phi_n(t))$ so that the total derivate kind of gives how much the vector of the inner derivatives increases the vector of the partial derivatives (or how much they point in the same direction).

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    $\begingroup$ This is not a definition, it's just the chain rule. $\endgroup$ – Crostul Apr 17 '18 at 13:32
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    $\begingroup$ Relevant thread: Intuitive reasoning behind the Chain Rule in multiple variables $\endgroup$ – littleO Apr 17 '18 at 13:50
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    $\begingroup$ I transcribed Terence Tao's intuitive explanation of the several variable chain rule here: math.meta.stackexchange.com/a/27339/40119 $\endgroup$ – littleO Apr 17 '18 at 13:57
  • $\begingroup$ Thabks @littleO I think I am almost there to fully grasping this. I think the only thing left is understanding why there is addition in the linear part of the approximation of the function. I mean, does the graphic of the function builds itself by adding the linear approximations in the different coordinates? OK, I think I see it now. $\endgroup$ – Nikola Apr 17 '18 at 14:16
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    $\begingroup$ One viewpoint to be aware of is that the nicest and most general version of the multivariable chain rule says that if $h(x) = f(g(x))$ then $h'(x) = f'(g(x)) g'(x)$ (under mild assumptions). Notice that on the right we are multiplying two matrices. The formula in your question is a special case of this matrix multiplication. $\endgroup$ – littleO Apr 17 '18 at 14:25
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It shows you the following based on the fact that the given expression is a dot-product:

  • imagine you run on a parametric curve on the surface given by $(x,f(x))$ through point $x^0$
  • if you run through $(x^0,f(x^0))$ in the direction of the gradient of $f$ at $x^0$, then you run in the direction of maximum increase on that surface
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