Differentiation

Since x=t+1      t = x-1     sustitute    y= 3(x-1)+2  = 3x-3+2 = 3x-1     then dy/dx = 3 

$\frac{dy}{dx};$  $x=t+1;$  $y=3t+2$

$\frac{d\times(3t+2)}{d\times(t+1)}$

$\frac{1\times(3t+2)}{1\times(t+1)}$

$\frac{3t+2}{t+1}$

x=t+1     t=x-1  $\frac{dt}{dx}=1$

y=3t+2    $\frac{dy}{dt}=3$

applying chain rule we have

$\frac{dy}{dx}=\frac{dy}{dt}*\frac{dt}{dx}=3$

Nice, fiora....!!!!