Taylor expansion with change of variables question. 
Find the Taylor polynomial of order 3 of
  $$f(x,y) = (x - 1)^{2} + \sin(\pi y) + x \ln(y)$$
  based at $(x,y) = (2,1)$.

So I'm really lazy and don't want to take the derivative of that, so let $u= x-2$ and $v= y-1$
$$f(x,y) = (u+1)^{2} + \sin(\pi v+\pi)) + u\ln(v+1) + 2 \ln(v+1)$$
$$\sin(\pi v+\pi)= -\sin(\pi v) + 0$$
$$f(x,y) = (u+1)^{2} - \sin(\pi v) + u\ln(v+1) + 2 \ln(v+1)$$
$$\ln(v+1) =   v -v^{2}/2 +v^{3}/3$$
$$\sin(\pi v) =  \pi [v-v^{3}/6]$$
$$(u+1)^{2} = 1 + 2(u) + 2u^{2}$$
$$1 + 2(u) + 2u^{2} - \pi [v-v^{3}/6] + u[v -v^{2}/2 +v^{3}/3] + 2[v -v^{2}/2 +v^{3}/3]$$
so all the 3rd order or less terms are
$$1 + 2(u) + 2u^{2} - \pi [v-v^{3}/6] + uv -uv^{2}/2 + 2[v -v^{2}/2 +v^{3}/3]$$
can i just sub back in now?
$$1 + 2(x-2) + 2(x-2)^{2} - \pi [y-1-(y-1)^{3}/6] + (x-2)(y-1) -(x-2)(y-1)^{2}/2 + 2[y-1 -(y-1)^{2}/2 +(y-1)^{3}/3]$$
 A: Write $x:=2+\xi$, $\ y:=1+\eta$. Then
$$\eqalign{\tilde f(\xi,\eta)&=(1+\xi)^2+\sin(\pi+\pi\eta)+(2+\xi)\log(1+\eta) \cr
&=1+2\xi+\xi^2-\pi\eta+{\pi^3\eta^3\over6}+(2+\xi)\left(\eta-{\eta^2\over2}+{\eta^3\over3}\right) +\ {\rm terms\ of\ order\ }\geq4\cr
&=1 +2\xi+(2-\pi)\eta+\xi^2+\xi\eta-\eta^2-{1\over2}\xi\eta^2+\left({\pi^3\over6}+{2\over3}\right)\eta^3+\ {\rm terms\ of\ order\ }\geq4\ .
\cr }$$ 
This is the Taylor expansion you are looking for, expressed in the "local variables" $\xi$, $\eta$ at $(2,1)$.
A: Your function is actually an easy one where the variables are not badly intertwined. And it only involves well-known Taylor series in one variable:
$$
\ln (1+u)=\sum_{n\geq 1}(-1)^{n+1}\frac{u^n}{n}\qquad\forall -1<u\leq 1
$$
and
$$
\sin v=\sum_{n\geq 0}(-1)^n\frac{v^{2n+1}}{(2n+1)!}\qquad \forall v\in\mathbb{R}.
$$
It follows that
$$
\ln y=\ln(1+(y-1))=\sum_{n\geq 1}(-1)^{n+1}\frac{(y-1)^n}{n}\qquad\forall 0<y\leq 2
$$
and
$$
\sin(\pi y)=\sin(\pi +\pi(y-1))=-\sin(\pi(y-1))
$$
$$
=\sum_{n\geq 0}(-1)^{n+1}\frac{\pi^{2n+1}(y-1)^{2n+1}}{(2n+1)!}\qquad \forall y\in\mathbb{R}.
$$
Finally
$$
(x-1)^2=(1+(x-2))^2=1+2(x-2)+(x-2)^2.
$$
Now you just have to plug these back in $f(x,y)$ and you are done.
You get:
$$
f(x,y)=(x-1)^2+\sin(\pi y)+((x-2)+2)\ln y= (x-1)^2+\sin (\pi y)+(x-2)\ln y+2\ln y
$$
$$
=1+2(x-2)+(x-2)^2-\pi(y-1)+\frac{\pi^3}{6}(y-1)^3+(x-2)(y-1)
$$
$$
-\frac{1}{2}(x-2)(y-1)^2+2(y-1)-(y-1)^2+\frac{2}{3}(y-1)^3.
$$
