Let $f:\mathbb{R}^2 \to \mathbb{R}, \ f(x,y) = \sin(e^x-y)$. What the quickest/a quick way to find its taylor polynomial of degree $3$ around the point $(0, \pi)$?
EDIT: More context... It is, of course, possible to just compute the derivatives but that seems kind of ugly. I'm looking for a more elegant way - possibly using the well-known Taylor expansion of $\sin(x)$.
Let $x = 0 + a, y = \pi + b$. We can try the "brute force" method, i.e. using the definition of the multivariate Taylor series
$$ f(a,\pi + b) = f(0,\pi) + f_x (0,\pi)a + f_y(0,\pi)b + f_{xx}(0,\pi)\frac{a^2}{2!} + f_{xy}(0,\pi)ab + f_{yy}(0,\pi)\frac{b^2}{2!} \\ + f_{xxx}(0,\pi)\frac{a^3}{3!} + f_{xxy}(0,\pi)\frac{a^2b}{2!} + f_{xyy}(0,\pi)\frac{ab^2}{2!} + f_{yyy}(0,\pi)\frac{b^3}{3!} $$
The derivatives shouldn't be difficult to compute \begin{align} f_x(x,y) &= e^x\cos(e^x-y) \\ f_y(x,y) &= -\cos(e^x-y) \\ f_{xx}(x,y) &= e^x\cos(e^x-y) - e^{2x}\sin(e^x-y) \\ f_{xy}(x,y) &= e^x\sin(e^x-y) \\ f_{yy}(x,y) &= -\sin(e^x-y) \\ f_{xxx}(x,y) &= e^x\cos(e^x-y) - 3e^{2x}\sin(e^x-y) - e^{3x}\cos(e^x-y) \\ f_{xxy}(x,y) &= e^x\sin(e^x-y) + e^{2x}\cos(e^x-y) \\ f_{xyy}(x,y) &= -e^x\cos(e^x-y) \\ f_{yyy}(x,y) &= \cos(e^x-y) \end{align}
There might be quicker ways, but this is the first thing that came to mind.