5672bf…: “From an autumn set”

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Compute $\dd [z]{t}$ for $z=x^2 y$ where $x$ and $y$ are functions of $t$ with $x=\sin (t)$ and $y=t^2+1$ in two different ways. First, leave your answer in terms of the gradient and the tangent vectors.

$\dd [z]{t}=\vector {\answer {2xy},\answer {x^2}}\dotp \vector {\dd [x]{t},\dd [y]{t}}$

And then, leave your answer in terms of $t$ alone.

$\dd [z]{t}=\answer {2 (t^2+1) \sin (t) \cos (t)+2t \sin ^2(t)}$