Find Taylor Polynomial of degree $2$ about (2,1) where $f(x,y) = x^2y^3$ , $(x,y) \in \mathbb{R}$

Question

Find Taylor Polynomial of degree $2$ about $(2,1)$ where $f(x,y) = x^2y^3$ $(x,y) \in \mathbb{R}$

My thoughts:

$D_xf= 2xy^3$ and $D_{xx}f= 2y^3$

$D_yf= x^23y^2$ and $D_{yy}f= x^26y$

$D_{xy} = 6xy^2 = D_{yx}$

$D_{xxy} = 6y^2$ and $D_{yyx} = 12xy$

$D_{xxx}=0$

But I'm not sure about the next steps... Any help is greatly appreciated!

Answer 1

The Taylor series of a two variable function about (a,b) is given by (using your notation):

$ f(x,y)= f(a,b)+[(x-a)D_{x}f(a,b)+(y-b)D_{y}f(a,b)]+\frac{1}{2!}[(x-b)^2D_{xx}f(a,b)+2(x-a)(y-b)D_{xy}f(a,b)+(y-b)^2D_{yy}f(a,b)]+... $

The phrase "Taylor polynomial of degree 2" refers to the degree of the $(x-a)$ and $(y-b)$ terms in front of each derivative.

You also need to evaluate each of your derivatives at (2,1) otherwise the result you get will not make any sense.