$\text { Let } w = \cos ( 4 x - 4 y ) \text {, where } x = t ^ { 9 } \text { and } y = 3 \text {. Find } \frac { d w } { d t } \text {. }$
A) $\frac { d w } { d t } = 12 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
B) $\frac { d w } { d t } = - 36 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
C) $\frac { d w } { d t } = 9 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
D) $\frac { d w } { d t } = 36 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
E) $\frac { d w } { d t } = - 12 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$

Question

$\text { Let } w = \cos ( 4 x - 4 y ) \text {, where } x = t ^ { 9 } \text { and } y = 3 \text {. Find } \frac { d w } { d t } \text {. }$ 
A) $\frac { d w } { d t } = 12 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
B) $\frac { d w } { d t } = - 36 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
C) $\frac { d w } { d t } = 9 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
D) $\frac { d w } { d t } = 36 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
E) $\frac { d w } { d t } = - 12 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$

Quizplus · Accepted Answer

The Answer of $\text { Let } w = \cos ( 4 x - 4 y ) \text {, where } x = t ^ { 9 } \text { and } y = 3 \text {. Find } \frac { d w } { d t } \text {. }$ 
A) $\frac { d w } { d t } = 12 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
B) $\frac { d w } { d t } = - 36 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
C) $\frac { d w } { d t } = 9 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
D) $\frac { d w } { d t } = 36 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$
E) $\frac { d w } { d t } = - 12 \sin \left( 4 t ^ { 9 } - 12 \right) t ^ { 8 }$

Let w=cos⁡(4x−4y), where x=t9 and y=3. Find dwdt. \text { Let } w = \cos ( 4 x - 4 y ) \text {, where } x = t ^ { 9 } \text { and } y = 3 \text {. Find } \frac { d w } { d t } \text {. } Let w=cos(4x−4y), where x=t9 and y=3. Find dtdw​.

$\text { Let } w = \cos ( 4 x - 4 y ) \text {, where } x = t ^ { 9 } \text { and } y = 3 \text {. Find } \frac { d w } { d t } \text {. }$