Answer
$$\frac{dy}{dt}=-\frac{130(5t+2)^{4}}{(3t-4)^{6}}$$
Work Step by Step
$$\frac{dy}{dt}=\frac{d}{dt}\Big(\frac{3t-4}{5t+2}\Big)^{-5}$$
Following the Chain Rule: $$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{d}{dt}\Big(\frac{3t-4}{5t+2}\Big)$$
$$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{(5t+2)\frac{d}{dt}(3t-4)-(3t-4)\frac{d}{dt}(5t+2)}{(5t+2)^2}$$
$$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{3(5t+2)-5(3t-4)}{(5t+2)^2}$$
$$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{15t+6-15t+20}{(5t+2)^2}$$
$$\frac{dy}{dt}=-5\Big(\frac{3t-4}{5t+2}\Big)^{-6}\frac{26}{(5t+2)^2}$$
$$\frac{dy}{dt}=-\frac{130(3t-4)^{-6}}{(5t+2)^{-4}}$$
$$\frac{dy}{dt}=-\frac{130(5t+2)^{4}}{(3t-4)^{6}}$$
