## University Calculus: Early Transcendentals (3rd Edition)

$$y'=-\frac{3t^2+6t+2}{t^2(t+1)^2(t+2)^2}$$
$$y=\frac{1}{t(t+1)(t+2)}$$ 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: $$\ln y=\ln\Big(\frac{1}{t(t+1)(t+2)}\Big)$$ Recall that $\ln(A\times B)=\ln A+\ln B$ and $\ln(A/B)=\ln A-\ln B$: $$\ln y=\ln1-\ln\Big(t(t+1)(t+2)\Big)=0-(\ln t+\ln(t+1)+\ln(t+2))$$ $$\ln y =-\ln t-\ln(t+1)-\ln(t+2)$$ 2) Then we take the derivative of both sides with respect to $t$ and remember that $(\ln t)'=1/t$: $$\frac{1}{y}\times y'=-\frac{1}{t}-\frac{1}{t+1}(t+1)'-\frac{1}{t+2}(t+2)'$$ $$\frac{y'}{y}=-\frac{1}{t}-\frac{1}{t+1}-\frac{1}{t+2}$$ $$\frac{y'}{y}=\frac{-(t+1)(t+2)-t(t+2)-t(t+1)}{t(t+1)(t+2)}$$ $$\frac{y'}{y}=\frac{-t^2-3t-2-t^2-2t-t^2-t}{t(t+1)(t+2)}$$ $$\frac{y'}{y}=\frac{-3t^2-6t-2}{t(t+1)(t+2)}=-\frac{3t^2+6t+2}{t(t+1)(t+2)}$$ 3) Solve for $y'$: $$y'=-\frac{3t^2+6t+2}{t(t+1)(t+2)}\times y$$ 4) Finally, substitute for $y=\frac{1}{t(t+1)(t+2)}$ $$y'=-\frac{3t^2+6t+2}{t(t+1)(t+2)}\times \frac{1}{t(t+1)(t+2)}=-\frac{3t^2+6t+2}{t^2(t+1)^2(t+2)^2}$$