# Total derivative not unique?

I probably did something wrong as I get two different total derivatives, but I don't see what. I use this definition: https://en.wikipedia.org/wiki/Total_derivative#The_total_derivative_as_a_linear_map

Let $f(x, y) = (x + y, x^2 + y^2, xy)$. Then $f(\xi + h) - f(\xi) = (h_1 + h_2, 2\xi_1h_1 + 2\xi_2h_2 + h_1^2 + h_2^2, \xi_1h_2 + \xi_2h_1 + h_1h_2) = (h_1 + h_2, 2\xi_1h_1 + 2\xi_2h_2, \xi_1h_2 + \xi_2h_1) + (0, ||h||^2, h_1h_2)$ but also $= (0, 2\xi_1h_1 + 2\xi_2h_2, \xi_1h_2 + \xi_2h_1) + (h_1 + h_2, ||h||^2, h_1h_2)$.

In both cases, if we fill it in the definition with the first vector the linear map aka total derivative and call the second $R(h)$ we get in both cases $0 < \frac{||R(h)||}{||h||} < \frac{||h||^2}{||h||} = ||h||$, so both tend to 0 as h tends to 0 by the squeeze theorem. But that would mean we have two different total differentials which is impossible. Where do I go wrong?

• Shouldn't it be a difference with the error term, not the sum? – Randall Sep 15 '17 at 11:56
• @Randall I don't think so? $||f(\xi + h) - f(\xi) - Df|| = ||Df + R(h) - Df|| = ||R(h)||$ – Pel de Pinda Sep 15 '17 at 11:59
• $h_1+h_2=O(\|h\|)$, not $o(\|h\|)$. – Gabriel Romon Sep 15 '17 at 12:02
• @GabrielRomon I don't know that notation. But my reasoning was as follows (let $v_i$ denote the i'th element from R(h)): $||R(h)|| = \sqrt{v_1^2 + ||h||^2 + v_3^2} < \sqrt{||h||^2} = ||h||^2$. – Pel de Pinda Sep 15 '17 at 12:05

## 2 Answers

Let $\|\cdot\|$ denote the $2$-norm in $\mathbb R^3$.

Note that $\|(h_1 + h_2, ||h||^2, h_1h_2)\|^2\geq (h_1+h_2)^2$.

Moreover $\frac{(h_1+h_2)^2}{h_1^2+h_2^2}$ does not go to $0$ as $\|h\|$ goes to $0$. Indeed, with $h_1=h_2=\frac 1m$, the ratio equals 2.

This means $\frac{\|(h_1 + h_2, ||h||^2, h_1h_2)\|}{\|h\|}$ does not go to $0$ as $\|h\|$ goes to $0$.

The total derivative is a special case of Fréchet derivative, which is uniquely defined. See this

Let $h=\pmatrix{\xi \\ \eta}.$ Write the increment of $f:$ $$f(x+\xi,\, y+\eta)-f(x,\,y) = \\ = \Big(x+\xi + y+\eta,\; (x+\xi)^2 + (y+\eta)^2,\; (x+\xi)(y+\eta)\Big) -(x + y,\; x^2 + y^2,\; xy) = \\ = (\xi + \eta, \; 2\xi x + \xi^2 + 2\eta y +\eta^2 , \; \xi y + \eta x + \xi\eta)= \\ =(\xi + \eta, \; 2\xi x + 2\eta y, \; \xi y + \eta x) + (0,\; \xi^2 +\eta^2,\; \xi\eta).$$ Thus the linear map, corresponding to the total derivative, can be represented as a matrix $$Df(x,\,y)= \pmatrix{1 & 1 \\ 2x & 2y \\ y & x}$$ since $$\left\Vert (0,\; \xi^2 + \eta^2,\; \xi\eta) \right\Vert \leqslant 2 \left\Vert h \right\Vert^2 = o(\left\Vert h \right\Vert), \quad \left\Vert h \right\Vert \to 0.$$