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So I've been asked a question: "Show the function $f(x,y) = x^2 - xy + 3y^2$ is differentiable at the point $(1,1)$ using the definition of differentiability." The question also mentions that it wants us to indicate the linear and little-o terms, and prove the little-o terms are in fact little-o terms.

I've been given the answer already (see below) and understand most of what is going on but I'm not sure how we go from line 3 to line 4 (line 4 being the one that starts with $f(1,1)$).

I understand what happens after line 4, it's just the jump from line 3 to line 4 that I don't understand, where to these terms ($o(h)$, $A h$, and $f(1,1)$) come from?

Thanks in advance My given answer

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1 Answer 1

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From line 3 to line 4:

$f(1,1)=3$

$h=(h_1,h_2)$ and $Ah=h_1+5h_2$

$h_1^2-h_1h_2+h_2^2=o(h)$

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  • $\begingroup$ ok, i get the f(1,1) part, he just subbed x=1 and y=1 into the original equation. For A, did he find the value of A just by seeing the values of the constants on h1 and h2 were 1 and 5? How did o(h) appear, is that just a general formula for little-o? $\endgroup$
    – Maximus
    Jan 28, 2021 at 15:18
  • $\begingroup$ @Maximus: $h_1+5h_2$ is the dot product of two vectors, $(1,5)$ and $(h_1,h_2)$. $\endgroup$
    – user9464
    Jan 28, 2021 at 15:20
  • $\begingroup$ @Maximus $h_1^2-h_1h_2+h_2^2=o(h)$ by definition of little o. Note particularly that $h=(h_1,h_2)$. $\endgroup$
    – user9464
    Jan 28, 2021 at 15:21
  • $\begingroup$ Ohhhh, ok thanks a lot, was actually pretty simple I just didn't look at it the right way. $\endgroup$
    – Maximus
    Jan 28, 2021 at 15:25
  • $\begingroup$ @Maximus You are welcome. :-) $\endgroup$
    – user9464
    Jan 28, 2021 at 15:26

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