Find the derivative of the function.
-$$s = \frac { t ^ { 8 } + 9 t + 4 } { t ^ { 2 } }$$
A) $\frac { d s } { d t } = t ^ { 5 } - \frac { 9 } { t ^ { 2 } } - \frac { 4 } { t ^ { 3 } }$
B) $\frac { d s } { d t } = 6 t ^ { 5 } + \frac { 9 } { t ^ { 2 } } + \frac { 8 } { t ^ { 3 } }$
C) $\frac { d s } { d t } = 6 t ^ { 5 } - \frac { 9 } { t ^ { 2 } } - \frac { 8 } { t ^ { 3 } }$
D) $\frac { d s } { d t } = 8 t ^ { 10 } + 9 t ^ { 2 } + 8 t$

Question

Quizplus · Accepted Answer

To find the derivative, we can use the quotient rule $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$. Applying this to $s=\frac{t^8+9t+4}{t^2}$, we get $s'=\frac{(t^8+9t+4)'(t^2)-(t^8+9t+4)(t^2)'}{(t^2)^2}$. Simplifying this gives $$s'=\frac{(8t^7+9)(t^2)-(t^8+9t+4)(2t)}{t^4}=\frac{16t^9-9t^8-18t-8}{t^4}.$$ Now, we can simplify this further by factoring out a $t^{-4}$, which gives $$s'=t^{-4}(16t^9-9t^8-18t-8)=16t^5-9t^{-2}-8t^{-3}.$$ This matches choice C.

Find the Derivative of the Function s=t8+9t+4t2s = \frac { t ^ { 8 } + 9 t + 4 } { t ^ { 2 } }s=t2t8+9t+4​

Find the Derivative of the Function $s = \frac { t ^ { 8 } + 9 t + 4 } { t ^ { 2 } }$