Find dy/dt.
-$$y = \left( e ^ { \cos ( t / 9 ) ) ^ { 4 } } \right.$$
A) $- \frac { 4 } { 9 } \left( e ^ { \sin ( t / 9 ) ) ^ { 3 } } \right.$
B) $\frac { 4 } { 9 } \sin \left( \frac { \mathrm { t } } { 9 } \right) \mathrm { e } ^ { 3 \cos ( \mathrm { t } / 9 ) }$
C) $- \frac { 4 } { 9 } \sin \left( \frac { \mathrm { t } } { 9 } \right) \mathrm { e } ^ { 4 \cos ( \mathrm { t } / 9 ) }$
D) $\frac { 4 } { 9 } \cos \left( \frac { t } { 9 } \right) \mathrm { e } ^ { 4 \cos ( \mathrm { t } / 9 ) }$

Question

Quizplus · Accepted Answer

To find $\frac{dy}{dt}$ for $y = \left( e ^ { \cos ( t / 9 ) ) ^ { 4 } } \right)$, we apply the chain rule. First, differentiate the outer function $e^{u}$ where $u = \cos(t/9)^4$, then multiply by the derivative of the inner function $\cos(t/9)^4$. The derivative of $e^u$ is $e^u$, and the derivative of $\cos(t/9)^4$ involves another application of the chain rule, giving $-\frac{4}{9}\sin(t/9)e^{4\cos(t/9)}$.

Find Dy/dt y=(ecos⁡(t/9))4y = \left( e ^ { \cos ( t / 9 ) ) ^ { 4 } } \right.y=(ecos(t/9))4

Find Dy/dt $y = \left( e ^ { \cos ( t / 9 ) ) ^ { 4 } } \right.$