Second order taylor series expansion of $\cos(x+2y)e^{2x}$ around $(2, -1)$ I'm asked to find the second order taylor series expansion of $\cos(x+2y)e^{2x}$ around $(2, -1)$. It's from an old exam of a professor who doesn't give solutions to them, so I don't know if my results are correct. So I woud like that you tell me if my results are correct, and if not, where I made a mistake. Thank you.

As first partial derivatives, I get $$f_x=-\sin(x+2y)e^{2x}+2\cos(x+2y)e^{2x}$$ $$f_y=-2\sin(x+2y)e^{2x}$$ As second partial derivatives, I get 
$$\begin{align}
f_{xx}&=-\cos(x+2y)e^{2x}-2\sin(x+2y)e^{2x}-2\sin(x+2y)e^{2x}+4\cos(x+2y)e^{2x}\\
&=3\cos(x+2y)e^{2x}-4\sin(x+2y)e^{2x}
\end{align}$$
$$f_{yy}=-4\cos(x+2y)e^{2x}$$ $$f_{xy}=f_{yx}=-2\cos(x+2y)e^{2x}-4\sin(x+2y)e^{2x}$$ 
So now, if I plug everything into the formula, I get 
$$e^{4}+2e^{4}(x-2)+\frac{1}{2!}(3e^{4}(x-2)^2-4e^{4}(x-2)(y+1)-4e^{4}(y+1)^2)=e^{4}(1+2(x-2)+\frac{1}{2!}(3(x-2)^2-4(x-2)(y+1)-4(y+1)^2))$$
Are my results correct ? I just proceeded as usual, using the common two-variables taylor series expansion. Thanks for your help !

It is correct. You can also obtain the expansion by using the Taylor series of $\sin, \cos,\exp$:
\begin{align}
\cos(x+2y)e^{2x} &= \cos((x-2) + 2(y+1))e^{4+2(x-2)}\\
&= e^{4}\big[\cos(x-2)\cos(2(y+1)) - \sin(x-2)\sin(2(y+1))\big]e^{2(x+2)}\\
&= e^{4}\left[\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(x-2)^{2n}\right)\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}2^{2n}(y+1)^{2n}\right)\right.\\
&- \left.\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(x-2)^{2n+1}\right)\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}2^{2n+1}(y+1)^{2n+1}\right)\right]\left(\sum_{n=0}^\infty \frac{2^n}{n!}(x-2)^{n}\right)\\
&= e^{4}\left[1+2(x-2) + \left(\frac{(-1)^1}{2!} - \frac{(-1)^1}{1!}\frac{2^1}{1!}\right)(x-2)^2 - \frac{(-1)^0}{0!}2^1\cdot \frac{2}{2!}(x-2)(y+1) + \frac{(-1)^1}{2!}2^2(y+2)^2 + \cdots\right]\\
&= e^4\left[1+2(x-2)+\frac32 e^{4}(x-2)^2-2e^{4}(x-2)(y+1)-2e^{4}(y+1)^2 + \cdots\right]
\end{align}
