Question

Use the Chain Rule to find dw/dt. (Enter your answer only in terms of t.)
`w = xe^(y/z), x = t^(4), y = 2 - t, z = 5 + 5t`

Answer 1

`{\Large \begin{array}{llll} w=t^4 e^{(2-t)/(5+5t)} \end{array}} \\\\[-0.35em] ~\dotfill\\\\ 4t^3e^{(2-t)/(5+5t)}~~ + ~~\stackrel{\textit{\LARGE product rule}}{t^4 \stackrel{\textit{\large chain rule}}{e^{(2-t)/(5+5t)}\cdot \stackrel{quotient~rule}{\cfrac{(-1)(5+5t)~~ - ~~(2-t)(5)}{(5+5t)^2}}}} \\\\\\ 4t^3e^{(2-t)/(5+5t)}~~ + ~~t^4 e^{(2-t)/(5+5t)}\cdot \cfrac{(-5-5t)~~ - ~~(10-5t)}{(5+5t)^2}`

`4t^3e^{(2-t)/(5+5t)}~~ + ~~t^4 e^{(2-t)/(5+5t)}\cdot \cfrac{-15}{(5+5t)^2} \implies t^3 e^{(2-t)/(5+5t)}\left[4- \cfrac{15t}{(5+5t)^2} \right] \\\\\\ t^3 e^{(2-t)/(5+5t)}\left[\cfrac{(100t^2+200t+100)-15t}{(5+5t)^2} \right]\implies t^3 e^{(2-t)/(5+5t)}\left[\cfrac{100t^2+185t+100}{(5+5t)^2} \right] \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} \cfrac{dw}{dt}=t^3 e^{(2-t)/(5+5t)}\left[\cfrac{100t^2+185t+100}{(5+5t)^2} \right] \end{array}}~\hfill`