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For the function $f(x,y)=x^2y$, I want to compute the directional derivative with direction $v=\langle1,1\rangle$ at point $(-1,-1)$. We end up getting $\begin{bmatrix}\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \end{bmatrix} \cdot\begin{bmatrix}2 \\ 1 \end{bmatrix}$, which is $\frac{3\sqrt{2}}{2}$.

However, thinking about it another way, $f(x,y)=x^2y$ cut along $v=\langle1,1\rangle$ is just $f(x,x)=x^3$. But when we take the derivative at the point mentioned above, we get $3$.

Clearly, the second argument is wrong, but why?

D.R.
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  • The functions $x^3$ $x^2y$ are only same in exactly the point $(1,1)$, in every neighbourhood of that point they will be at least a little different and partial derivatives used in the directional derivative concern these tiny neighbourhoods. – mathreadler May 14 '17 at 16:21

1 Answers1

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The directional derivative gives you the change of $f$ along $v$ per unit length moved in that direction.

In your case when you take $$\frac{d}{dx}x^3=3x^2$$ you are finding the change of $f$ along $v$ per unit change in $x$.

The two quantities are obviously not the same and differ by a factor of $\sqrt{2}$.

The directional derivative should be smaller because to have unit change in $x$, you need to move a distance of $\sqrt{2}$ in $v$.

Hence it should be $$3\times \frac{1}{\sqrt{2}}$$

velut luna
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