Show differentiability at a point then find differential where $f(x,y) = (x^2, xy+y^2)$

Question

Show differentiability at a point then find differential where

$f(x,y) = (x^2, xy+y^2)$

Want to show that $f(x,y) = (x^2, xy+y^2)$ is differentiable at $(a,b)$ and then calculate the differential d$f(a,b)$

My Thoughts

$f_x(x,y) = (2x, y)$ and $f_{xx}(x,y) = (2, 0)$

$f_y(x,y) = (0, x+2y)$ and $f_{yy}(x,y) = (0, 2)$

$f_{xy}(x,y) = (0,1) = f_{yx}(x,y)$

At point $(a,b)$ we get that $f_x(a,b) = (2a,b)$ and $f_y(a,b) = (0,a+2b)$

I'm not sure how to proceed from here. Am I even on the right track? I am really struggling with differentials.

Any help is greatly appreciated! Thanks!

Answer 1

The function is a polinomial, so all partials exist and are continuous everywhere, therefore, your function is differentiable. The Fréchet-differential's matrix in the standard basis is $$ DF|_{(a,b)}=\begin{pmatrix} \left.\frac{\partial f_1}{\partial x}\right|_{(a,b)} & \left.\frac{\partial f_1}{\partial y}\right|_{(a,b)} \\ \left.\frac{\partial f_2}{\partial x}\right|_{(a,b)} & \left.\frac{\partial f_2}{\partial y}\right|_{(a,b)} \end{pmatrix}. $$

Answer 2

$$\newcommand{\rac}[1]{\left\langle#1\right\rangle} f(x,y)=\rac{x^2,xy+y^2}\\ f_x(x,y)=\rac{2x,y}\\ f_y(x,y)=\rac{0,x+2y}\\ f_{xy}(x,y)=\rac{0,1}\\ f_{yx}(x,y)=\rac{0,1}\\ f_{xy}=f_{yx}$$

Answer 3

If you are just beginning with differentials, you can come back to the definition:

$f(a+h_1,b+h_2)=(a^2+2ah_1+h_1^2,ab+ah_2+bh_1+h_1h_2+b^2+2bh_2+h_2^2)$ $=(a^2,ab+b^2)+(2ah_1,ah_2+bh_1+2bh_2)+(h_1^2,h_1h_2+h_2^2)$

Now you can check that $(h_1^2,h_1h_2+h_2^2)$ is composed by second order terms in $h_1, h_2$, and that your differential will be

$(h_1,h_2)\mapsto (2ah_1,ah_2+bh_1+2bh_2)$

so that you have

$f(a+h_1,b+h_2)=f(a,b)+df_{(a,b)}(h_1,h_2)+\|h_1,h_2\|\varepsilon(h_1,h_2)$

with $\lim\limits_{(h_1,h_2)\to(0,0)}\varepsilon(h_1,h_2)=0$