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....!!!!