Given $y = \frac{\sin(x)}{e^x}$, we need to find the first, second, third, and fourth derivatives of $y$ with respect to $x$.

Step 1: Find the first derivative $\frac{dy}{dx}$

We will use the quotient rule for differentiation, which states: $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ where $f(x) = \sin(x)$ and $g(x) = e^x$.

1. $f'(x) = \cos(x)$
2. $g'(x) = e^x$

Now, apply the quotient rule: $\frac{dy}{dx} = \frac{\cos(x) \cdot e^x - \sin(x) \cdot e^x}{(e^x)^2}$ Simplify the expression: $\frac{dy}{dx} = \frac{e^x(\cos(x) - \sin(x))}{e^{2x}} = \frac{\cos(x) - \sin(x)}{e^x}$

Step 2: Find the second derivative $\frac{d^2y}{dx^2}$

Now, differentiate $\frac{dy}{dx} = \frac{\cos(x) - \sin(x)}{e^x}$ using the quotient rule again.

Let $u(x) = \cos(x) - \sin(x)$ and $v(x) = e^x$.

1. $u'(x) = -\sin(x) - \cos(x)$
2. $v'(x) = e^x$

Using the quotient rule: $\frac{d^2y}{dx^2} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$ Substituting in the values: $\frac{d^2y}{dx^2} = \frac{(-\sin(x) - \cos(x)) \cdot e^x - (\cos(x) - \sin(x)) \cdot e^x}{e^{2x}}$ Simplify the expression: $\frac{d^2y}{dx^2} = \frac{e^x(-\sin(x) - \cos(x) - \cos(x) + \sin(x))}{e^{2x}} = \frac{-2\cos(x)}{e^x}$

Step 3: Find the third derivative $\frac{d^3y}{dx^3}$

Differentiate $\frac{d^2y}{dx^2} = \frac{-2\cos(x)}{e^x}$.

Using the quotient rule again, let $u(x) = -2\cos(x)$ and $v(x) = e^x$.

1. $u'(x) = 2\sin(x)$
2. $v'(x) = e^x$

$\frac{d^3y}{dx^3} = \frac{2\sin(x) \cdot e^x - (-2\cos(x)) \cdot e^x}{e^{2x}}$ Simplify the expression: $\frac{d^3y}{dx^3} = \frac{e^x(2\sin(x) + 2\cos(x))}{e^{2x}} = \frac{2(\sin(x) + \cos(x))}{e^x}$

Step 4: Find the fourth derivative $\frac{d^4y}{dx^4}$

Differentiate $\frac{d^3y}{dx^3} = \frac{2(\sin(x) + \cos(x))}{e^x}$.

Using the quotient rule again, let $u(x) = 2(\sin(x) + \cos(x))$ and $v(x) = e^x$.

1. $u'(x) = 2(\cos(x) - \sin(x))$
2. $v'(x) = e^x$

$\frac{d^4y}{dx^4} = \frac{2(\cos(x) - \sin(x)) \cdot e^x - 2(\sin(x) + \cos(x)) \cdot e^x}{e^{2x}}$ Simplify the expression: $\frac{d^4y}{dx^4} = \frac{2e^x(\cos(x) - \sin(x) - \sin(x) - \cos(x))}{e^{2x}} = \frac{-4\sin(x)}{e^x}$

Summary of Derivatives:

1. $\frac{dy}{dx} = \frac{\cos(x) - \sin(x)}{e^x}$
2. $\frac{d^2y}{dx^2} = \frac{-2\cos(x)}{e^x}$
3. $\frac{d^3y}{dx^3} = \frac{2(\sin(x) + \cos(x))}{e^x}$
4. $\frac{d^4y}{dx^4} = \frac{-4\sin(x)}{e^x}$

Would you like any further explanation or details on any part of the process?

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5 Related Questions:

1. How would the derivatives change if $y = \frac{\cos(x)}{e^x}$?
2. Can you solve for the nth derivative of $y = \frac{\sin(x)}{e^x}$?
3. What is the behavior of the function $y = \frac{\sin(x)}{e^x}$ as $x$ approaches infinity?
4. How do the derivatives of exponential functions generally behave?
5. What role does the chain rule play in more complex functions?

Tip: Always check if simplifications can be made at each step of differentiation to avoid errors and make the final expression more manageable.