Taylor Expansion of a Trig Function

Question

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.

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$ \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)

Answer 1

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

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

Answer 2

If that is indeed the calculation that you are both trying to execute (perhaps this is where the misunderstanding lies?), then you are right.