# Why the gradient vector gives the direction of maximum increase of a function?

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.

• 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...? – Pedro Tamaroff Sep 2 '16 at 1:42
• 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 – reuns Sep 2 '16 at 1:46

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.
• 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. – Matthew Leingang Sep 2 '16 at 1:49