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Why the gradient vector gives the direction of maximum increase of a function? In context of multivariable functions, $$f: \Bbb{R}^2\to\Bbb{R} $$

Y know that the gradient vector is defined as $$\nabla f(x,y) = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$$

And I understand that the partial derivatives gives the increase value in the directions of i and j versor respectively. But, why the gradient vector, compound of these two values gives the direction of maximum increase? Why can't be another vector or direction which gives that? Thank you.

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    $\begingroup$ The directional derivative of $f$ in the direction $v$ at point $p$ is the dot product $\langle \nabla f(p),v\rangle$. By Cauchy--Schwarz this is maximed when...? $\endgroup$ – Pedro Tamaroff Sep 2 '16 at 1:42
  • $\begingroup$ if you don't like Cauchy-Schwarz : check that the gradient is $|C|(1,0)$ when the direction of maximum increase is $(1,0)$. then use the linearity of the gradient with respect to a rotation of the coordinates $\endgroup$ – reuns Sep 2 '16 at 1:46
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The directional derivative $D_v = \nabla f_p \cdot v = \|\nabla f_p\| \cos \theta_{\nabla f_p,v}$ and since $-1 \leq \cos t \leq 1$ then the derivative is maximal with value $\|\nabla f_p\|$ i.e in the direction of the gradient.

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    $\begingroup$ To add a few steps between: The maximum value of $D_vf$ at $p$ happens when $\cos\theta = 1$. This happens when $\theta = 0$, that is, when $\nabla f_p$ and $v$ are pointing in the same direciton. $\endgroup$ – Matthew Leingang Sep 2 '16 at 1:49
  • $\begingroup$ Yes, I understand that the directional derivative is maximized when goes in the same direction of the gradient vector. But I can't understand why the gradient itself gives the maximun increase direction, why can't be another vector? Why the gradient have this particularity? $\endgroup$ – DN_Euler Sep 2 '16 at 2:32
  • $\begingroup$ By definition the directional derivative depends only on the gradient and the directional vector. There are no other vectors to consider. $\endgroup$ – Faraad Armwood Sep 2 '16 at 2:52

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