设函数y=y(x)y = y(x)y=y(x)由{x=2t−1y=t2+2\begin{cases}x = 2t - 1 \\ y = t^2 + 2\end{cases}{x=2t−1y=t2+2所确定,则dydx=\dfrac{dy}{dx}=dxdy=( ).
(A)2t
(B)t
(C)2
(D)t2t^2t2
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