Given that y= sinx / e^x find dy/dx then find d^2y / d^2x then find d^3y / d^3x then find d^4y / d^4x

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.

Learn how to differentiate the function y = sin(x) / e^x step-by-step with detailed solutions for first, second, third, and fourth derivatives. Understand the application of the quotient rule and its use in differentiating exponential functions. Suitable for college-level mathematics students.

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