Solution
= Solution
The <product rule> and <chain rule> give $y'=\sin(x^2)+2x^2\cos(x^2)$. At zero the slope is zero and the <curve> passes through $(0,1)$. Consequently the <tangent line> is
$$
\boxed{y=1}.
$$
= Solution
The <product rule> and <chain rule> give $y'=\sin(x^2)+2x^2\cos(x^2)$. At zero the slope is zero and the <curve> passes through $(0,1)$. Consequently the <tangent line> is
$$
\boxed{y=1}.
$$