The gradient vector of a function of two variables $f(x,y)$ is $$\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right).$$ and usually, people say that the gradient vector tells you what is the direction you should take to increase the value of the function faster, but I can't relate this with taking the derivatives of the function, I mean what is the relation between the derivatives of $f$ and the direction that increases the function faster?
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See this answer I wrote. It's an application in machine learning but it more or less captures the essnece of your quesiton – Clemens Bartholdy May 28 '21 at 00:03
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1strictly speaking the gradient is not a vector, instead is a covector field – janmarqz May 28 '21 at 01:01
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@janmarqz I wonder when elementary calculus / analysis classes will finally stop teaching this misconception. – Charles Hudgins May 28 '21 at 01:28
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@janmarqz: The differential $df$ is a covector field. The gradient $\nabla f$ is the vector field obtained by raising the index on $df$ using the metric. – Hans Lundmark May 28 '21 at 06:12
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Consider the direction $v = (a,b)$, where $\|v\| = 1$.
Then the directional derivative in the direction $v$ is given by: \begin{align*} D_{v}f(x,y) = f'(x,y)v = \|\nabla f\|\|v\|\cos(\theta) = \|\nabla f\|\cos(\theta) \end{align*} where $\theta$ is the angle between $\nabla f$ and $v$.
It attains its maximum value when $\theta = 0$.
So $v$ which maximizes the directional derivative is in the same direction as $\nabla f$.
Hopefully this helps!
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