## Thomas' Calculus 13th Edition

$\dfrac{1}{2}$
Here, we have $\dfrac{dx}{dt}=2(x-t)$ and $\dfrac{dy}{dt}=(e^t)(t+1)$ Here, $x-t=1$ when $t=0$ $\implies \dfrac{dx}{dt}=2$ Also, $\dfrac{dx}{dt}(t=0)=1$ Thus, Slope: $\dfrac{dy}{dx}=\dfrac{(e^{0})(0+1)}{2(1)}=\dfrac{1}{2}$