Answer
$$y'=3t^2+6t+2$$
Work Step by Step
$$y=t(t+1)(t+2)$$
Logarithmic differentiation helps to simplify the differentiation of otherwise complex functions by using the algebraic properties of logarithm. In detail,
1) Take the natural logarithm of both sides and simplify:
$$\ln y=\ln\Big(t(t+1)(t+2)\Big)$$
Recall that $\ln(A\times B)=\ln A+\ln B$: $$\ln y=\ln t+\ln(t+1)+\ln(t+2)$$
2) Then we take the derivative of both sides with respect to $t$ and remember that $(\ln t)'=1/t$:
$$\frac{1}{y}\times y'=\frac{1}{t}+\frac{1}{t+1}(t+1)'+\frac{1}{t+2}(t+2)'$$ $$\frac{y'}{y}=\frac{1}{t}+\frac{1}{t+1}+\frac{1}{t+2}$$ $$\frac{y'}{y}=\frac{(t+1)(t+2)+t(t+2)+t(t+1)}{t(t+1)(t+2)}$$ $$\frac{y'}{y}=\frac{(t^2+3t+2)+(t^2+2t)+(t^2+t)}{t(t+1)(t+2)}$$ $$\frac{y'}{y}=\frac{3t^2+6t+2}{t(t+1)(t+2)}$$
3) Solve for $y'$: $$y'=\frac{3t^2+6t+2}{t(t+1)(t+2)}\times y$$
4) Finally, substitute for $y=t(t+1)(t+2)$
$$y'=\frac{3t^2+6t+2}{t(t+1)(t+2)}\times t(t+1)(t+2)=3t^2+6t+2$$
