Taylor polynomial about the origin Find the 3rd degree Taylor polynomial about the origin of $$f(x,y)=\sin (x)\ln(1+y)$$

So I used this formula to calculate it $$p=f(0,0)+(f_x(0,0)x+f_y(0,0)y)+(\frac{1}{2}f_{xx}(0,0)x^2+f_{xy}(0,0)xy+\frac{1}{2}f_{yy}(0,0)y^2)+(\frac{1}{6}f_{xxx}(0,0)x^3+\frac{1}{2}f_{xxy}(0,0)x^2y+\frac{1}{2}f_{xyy}(0,0)xy^2+\frac{1}{6}f_{yyy}(0,0)y^3)$$
I get $x(\ln(1)+y-\frac{\ln(1)x^2}{6}-\frac{y^2}{2})$
But as you can see, this is a very tedious task (especially if I have to do this on my midterm). 
There exists a Taylor series for $\sin(x)$ and $\ln(1+y)$.
If I only keep the terms with degree $\le 3$, I have $$\sin(x)\ln(1+y)=(x-\frac{x^3}{3!})(y-\frac{y^2}{2}+\frac{y^3}{3}) \\=xy-\frac{xy^2}{2}$$
(I multiply the two and remove terms with degree > 3 from the answer)  
The two polynomials are different. Is the second method even a valid way to determine Taylor polynomial?

Since your function is a product of the form $f(x)g(y)$ this is valid.  If you had a term like $e^{-xy}$ instead, then one would have to use the full Taylor series formula that you mentioned above.
