Find $\frac { d z } { d t }$ for $z = \frac { x } { y } , x = t ^ { 2 } s ^ { 2 } , y = s e ^ { t }$
A) $- 2 s e ^ { - t }$
B) $2 s t e ^ { - t } - s t ^ { 2 } e ^ { - t }$
C) $2 s t e ^ { - t } - s t ^ { 2 } e ^ { - 2 t }$
D) $2 t e ^ { - t } - t ^ { 2 } e ^ { - t }$

Question

Quizplus · Accepted Answer

To find $\frac{dz}{dt}$ for $z = \frac{x}{y}$, where $x = t^2s^2$ and $y = se^t$, we use the quotient rule for differentiation: $\frac{d}{dt}\left(\frac{x}{y}\right) = \frac{y\frac{dx}{dt} - x\frac{dy}{dt}}{y^2}$. First, differentiate $x = t^2s^2$ with respect to $t$ to get $\frac{dx}{dt} = 2t s^2$. Then, differentiate $y = se^t$ with respect to $t$ to get $\frac{dy}{dt} = se^t$. Substituting these into the quotient rule gives $\frac{dz}{dt} = \frac{se^t(2ts^2) - t^2s^2(se^t)}{(se^t)^2} = 2st e^{-t} - st^2e^{-t}$.

Find dzdt\frac { d z } { d t }dtdz​ For z=xy,x=t2s2,y=setz = \frac { x } { y } , x = t ^ { 2 } s ^ { 2 } , y = s e ^ { t }z=yx​,x=t2s2,y=set

Find $\frac { d z } { d t }$ For $z = \frac { x } { y } , x = t ^ { 2 } s ^ { 2 } , y = s e ^ { t }$