Taylor Polynomial of Degree $2$ Question

Question

For the function $f(x,y,z)=cos(ye^x)sin(x)$, find the corrrsponding Taylor polynomial of degree $2$ at $(x,y)=(0,\frac{\pi}{2})$.

I tried to compute all first second order partial derivatives at $(x,y)=(0,\frac{\pi}{2})$, but all gave 0. Feeding this into $f(x,y)=f_x(0,0)(x-0)+f_y(0,0)(y-0)+\frac{1}{2}(f_{xx}(0,0)(x-0)^2+2f_{xy}(0,0)(x-0)(y-0)+f_{yy}(0,0)(y-0)^2$, the answer is 0, but this is not the correct solution. Please help.

Answer 1

You got the wrong result because your computations are wrong. Although $\frac{\partial^2f}{\partial y^2}\left(0,\frac\pi2\right)$ is indeed $0$, you have $\frac{\partial^2f}{\partial x^2}\left(0,\frac\pi2\right)=-\pi$ and $\frac{\partial^2f}{\partial x\partial y}\left(0,\frac\pi2\right)=-1$.