Let $y=x^3-8x+7\text{ and }x=f\left(t\right)\text{. }$ and $.\text{ If }\frac{dy}{dt}=2,$$x=3\text{ at }t=0\text{, then the value of }\left(\frac{dx}{dt}\right)\text{ at }t=0=\_\_\_\_\_\_$

$y=x^3-8x+7\Rightarrow \frac{dy}{dx}=3x^2-8$

$\text{ It is given that when }\mathrm{t}=0;x=3$

$\text{ when }t=0;\frac{dy}{dx}=3\times 3^2-8=19$

$\frac{dy}{dx}=\frac{(dy/dt)}{(dx/dt)}......\left(1\right)$

$\text{ when }t=0;\frac{dy}{dx}=19\text{ and }$

$\frac{dy}{dt}=2\Rightarrow \frac{dx}{dt}=2\left(\frac{1}{19}\right)$

$\therefore \frac{dx}{dt}=\frac{2}{19}$