Find y .
-$y = \sin \left( 4 x ^ { 2 } e ^ { x } \right)$
A) $\left( x ^ { 2 } + 4 x + 2 \right) e ^ { x } \cos \left( 4 x ^ { 2 } e ^ { x } \right) + 4 x e ^ { x } \left( x ^ { 3 } + 4 x ^ { 2 } + 4 x \right) \cos \left( 4 x ^ { 2 } e ^ { x } \right)$
B) $4 \left( x ^ { 2 } + 4 x + 2 \right) e ^ { x } \cos \left( 4 x ^ { 2 } e ^ { x } \right) - 16 x e ^ { 2 x } \left( x ^ { 3 } + 4 x ^ { 2 } + 4 x \right) \sin \left( 4 x ^ { 2 } e ^ { x } \right)$
C) $8 x e ^ { x } \cos \left( 4 x ^ { 2 } e ^ { x } \right) + 4 x ^ { 2 } e ^ { x } \cos \left( 4 x ^ { 2 } e ^ { x } \right)$
D) $- 8 e ^ { x } \sin \left( 4 x ^ { 2 } e ^ { x } \right) - 8 x e ^ { x } \sin \left( 4 x ^ { 2 } e ^ { X } \right)$

Question

Quizplus · Accepted Answer

This is a chain rule problem, where we need to take the derivative of the outer function first, and then multiply it by the derivative of the inner function. Applying the chain rule, we get:
$$\frac{dy}{dx} = \cos \left( 4 x ^ { 2 } e ^ { x } \right) \cdot \frac{d}{dx} \left( 4 x ^ { 2 } e ^ { x } \right)$$
Using the product rule, we can simplify this to:
$$\frac{dy}{dx} = \cos \left( 4 x ^ { 2 } e ^ { x } \right) \cdot \left( 8 x e ^ { x } + 4 x ^ { 2 } e ^ { x } \right)$$
Simplifying further, we get:
$$\frac{dy}{dx} = 4 \left( x ^ { 2 } + 4 x + 2 \right) e ^ { x } \cos \left( 4 x ^ { 2 } e ^ { x } \right) - 16 x e ^ { 2 x } \left( x ^ { 3 } + 4 x ^ { 2 } + 4 x \right) \sin \left( 4 x ^ { 2 } e ^ { x } \right)$$
Therefore, the correct answer is B.

Find Y - y=sin⁡(4x2ex)y = \sin \left( 4 x ^ { 2 } e ^ { x } \right)y=sin(4x2ex)

Find Y
- $y = \sin \left( 4 x ^ { 2 } e ^ { x } \right)$