Calculus: Early Transcendentals 8th Edition

$f'(t)=-\frac{2t+3}{3t^{2/3}(t-3)^2}$
$f(t)=\frac{\sqrt[3] t}{t-3}$ We will differentiate using the quotient rule $f'(t)=\frac{(t-3)(\sqrt[3] t)'-(\sqrt[3] t)(t-3)'}{(t-3)^{2}}$ Rewrite the square root using fractional power and continue $f'(t)=\frac{(t-3)(t^{1/3})'-(t^{1/3})(t-3)'}{(t-3)^{2}}$ $(t^{1/3})'= \frac{1}{3}t^{1/3-1}= \frac{1}{3}t^{-2/3}$ $(t-3)'=1$ $f'(t)= \frac{(t-3)\frac{1}{3}t^{-2/3}-(t^{1/3})(1)}{(t-3)^{2}}$ Evaluate products $f'(t)=\frac{\frac{t-3}{3t^2/3}-t^{1/3}}{(t-3)^2}$ Simplify $f'(t)=-\frac{2t+3}{3t^{2/3}(t-3)^2}$