I've got to prove the existence and uniqueness of a solution of $y'(x)=sin(x^2y^2)$ with $y(0)=1$ in the rectangle $R=\{(x,y):|x|\leq 1, |y-1|\leq1\}$ in the intervall $[-1,1]$.
So after showing that $f(x,y)=sin(x^2y^2)$ is continuously differentiable with $f_y(x,y)=2yx^2cos(x^2y^2)$, we can write : $|f_y(x,y)| \leq 2x^2y$ as $|cos(z)| \leq 1$. Now my tutor has stated that it follows $|f_y(x,y)| \leq 2x^2y\leq 2\cdot 1 \cdot 4$.. Why he has written these numbers? I only come to $2\cdot1\cdot 1$.
for $x$ is given $$|x|\le 1$$ so $$x^2\le 1$$ for $y$ is given $$|y-1|\le 1$$ and this is equivalent with $$0\le y\le 2$$