<s>[INST] Taylor Expansion of a Trig Function Did a problem in a lecture today with Taylor Expansions. 
The question is: 
Find the Taylor Expansion of 
$
z=\sin(x^2+y^2)
$
To the second order at the point (1, -1)
I am happy with the solution until she gets to 
$
\frac{d^2(f)}{d(y)^2}(1,-1)=-4\sin(2)=-3,637
$
According to my own calculations,
$
\frac{d^2(f)}{d(y)^2}(1,-1)=2\cos(2)-4\sin(2)=-4,469
$
She checked her answers and insists that she is correct. I am confused as to where that $2\cos(2)$ went. If someone could explain it to me that would be awesome.

$
\frac {df}{dy}=2y\cos (x^2+y^2)
$
And
$
\frac {d^2(f)}{d(y)^2}=2\cos(x^2+y^2)-4(y)^2 \sin(x^2+y^2)
$
(Using product rule)
 [/INST] Your computations are fine. We have
$$
f(x,y):=\sin(x^2+y^2)
$$ giving by the chain rule
$$
\frac {df}{dy}(x,y)=2y\cos (x^2+y^2), \qquad \frac {d^2f}{d^2y}(x,y)=2\cos(x^2+y^2)-4y^2 \sin(x^2+y^2)
$$ and

$$
\left.\frac {d^2f}{d^2y}(x,y)\right|_{(1,-1)}=2\cos(2)-4 \sin(2)=-4.469483380\cdots.
$$

</s>