Find dy/dt.
-$$y = t ^ { 4 } \left( t ^ { 6 } + 8 \right) ^ { 6 }$$
A) $4 t ^ { 3 } \left( t ^ { 6 } + 8 \right) ^ { 5 } \left( 36 t ^ { 6 } + 8 \right)$
B) $t ^ { 3 } \left( t ^ { 6 } + 8 \right) ^ { 5 } \left( 40 t ^ { 6 } + 32 \right)$
C) $144 t ^ { 22 } \left( t ^ { 6 } + 8 \right) ^ { 5 }$
D) $t ^ { 4 } \left( t ^ { 6 } + 8 \right) ^ { 5 } \left( 40 t ^ { 5 } + 32 \right)$

Question

Quizplus · Accepted Answer

Using the chain rule, we have:
$$\frac { dy } { dt } = \frac { d } { dt } \left( t ^ { 4 } \left( t ^ { 6 } + 8 \right) ^ { 6 } \right) = 4t ^ { 3 } \left( t ^ { 6 } + 8 \right) ^ { 6 } + 6t ^ { 4 } \cdot 6 \left( t ^ { 6 } + 8 \right) ^ { 5 } \cdot 6t ^ { 5 }$$
Simplifying, we get:
$$\frac { dy } { dt } = t^3(t^6+8)^5(40t^6+32) = t^3\left(t^6+8\right)^5\cdot4\left(10t^6+8\right)$$
Therefore, the answer is choice B.

Find Dy/dt y=t4(t6+8)6y = t ^ { 4 } \left( t ^ { 6 } + 8 \right) ^ { 6 }y=t4(t6+8)6

Find Dy/dt $y = t ^ { 4 } \left( t ^ { 6 } + 8 \right) ^ { 6 }$