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I'm currently trying to understand the proof of the directional derivative of a multivariate function $f(x,y)$ at the point $(x_0,y_0)$ along the vector $\vec u \langle a,b \rangle $.

$$D_\vec u f(x_0,y_0) = \lim_{h\to0} \dfrac{f(x_0 + ah, y_0 + bh) - f(x_0, y_0)}{h}$$

Then, it says let's put:

$$ g(h) = f(x_0 + ah, y_0 + bh) $$

How can we change a function of 2 variables to a function of only one variable, which is a totally different one. Is it really valid? How is this process called, I couldn't find the exact name to search for more details. Thank you.

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Pick a function, $f(x,y)=x^2+y^2$, a point, $P(x_0,y_0)=(1,2)$ and a direction $\vec u=(3,4)$, then calculate the derivative of $f$ at $P$ along $\vec u$

$$D_\vec u f(x_0,y_0) = \lim_{h\to0} \dfrac{f(x_0 + ah, y_0 + bh) - f(x_0, y_0)}{h}$$

$$D_{(3,4)} f(1,2) = \lim_{h\to0} \dfrac{(1 + 3h)^2+(2 + 4h)^2 - (1^2+2^2)}{h}$$

Clearly $f(x_0 + ah, y_0 + bh)=(1 + 3h)^2+(2 + 4h)^2$ is a one variable function in $h$. We can call it $g$, so is

$g(h)=(1 + 3h)^2+(2 + 4h)^2=25h^2+22h+5$

$g$ is the value of $f$ fixing a point and moving along a fixed direction. The procedure is part, as you see, of the construction of the incremental quotient, the very basis of the differential calculus and, to me, a useful starting point for the intuition about the subject.

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  • $\begingroup$ It is much more clear said this way. Thank you! $\endgroup$ – Ryan B. Nov 23 '18 at 16:30
  • $\begingroup$ You are welcome :) $\endgroup$ – Rafa Budría Nov 23 '18 at 16:31

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