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I am stuck with the following problem:

Consider the function $f(x,y)=\frac{x^2}{y^2}$; where $(x,y)\in [1/2,3/2]\times[1/2,3/2]$.Then what is the derivative of $f$ at $(1,1)$along the direction $(1,1)$?

Can someone explain it ? Thanks in advance for your time.

WHAT I TRIED:

I would like to find all directional derivatives of the function $$f(x,y) = x^2y^{-2} , $$ (where $ (x,y) \in \mathbb{R}^2 $), in the point $(0,0)$. I tried to do this by calculating $$\nabla f(x,y) = f_1 (x,y) e_1 + f_2 (x,y) e_2 $$, where $e_n$ is the $n$'th unit vector and $f_n$ is the partial derivative with respect to the $n$'th variable. I found that $f_1 (x,y) = 2xy^{-2}$, and that $f_2 (x,y) = -2x^2y^{-3} $.

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The function $f $ has gradient $\nabla f = (2x/y^2, -2x^2/y^3), $ so at the point $(1,1) $ the gradient is $(2,-2) $. The directional derivative is simply the dot product of the gradient with a normalized direction vector: $(2,-2)\cdot (1,1)/\|(1,1)\|= 0. $

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