How can I approach this exercise using the Chain Rule?
Let $f: \mathbb{R}^2 \to \mathbb{R}$, $C^2$ function whose second degree Taylor polynomial centered at (1,0) is $P(x,y) = 4 + x + xy + \frac{y^2}{2}$.
Let $g: \mathbb{R}^2 \to \mathbb{R}, g(x,y)= e^{f(x,y) - 5} \sin(y)$. Find the second degree Taylor polynomial of $g$ at (1,0).
I'd like to break down $g$ as a composition $g(x,y) = h(f(x,y))$. Can I "redefine" $f'(x,y)=(f(x,y),y)$ such that $D_g = D_h \cdot D_{f'}$? Then $D_{f'}$ would be $\begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ 0 & 1 \end{bmatrix}$.
Is there any simple way to find $D^2 g$ at (1,0)?
HINT: First, you know the Taylor polynomials at $0$ of $\sin(y)$ and $e^u$. Note that $f(1,0)=5$, so $f(x,y)-5 = 0$ at $(1,0)$. I suggest you substitute $u=f(x,y)-5$. Then the Taylor polynomial of the product is obtained by multiplying the Taylor polynomials and dropping the terms of too high a degree.
You can think of this in terms of a composition of functions, of course: Let $G(u,v) = e^u\sin v$, and consider $G(f(x,y)-5,y)$.