Find the second derivatives of the following functions :
`e^(-x)cosx`

Answer

To find the second derivative of the function \( y = e^{-x} \cos x \), we will follow these steps:

### Step 1: Differentiate the function to find the first derivative

We will use the product rule for differentiation, which states that if \( y = u \cdot v \), then:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Here, let \( u = e^{-x} \) and \( v = \cos x \).

Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

- \( \frac{du}{dx} = -e^{-x} \) (derivative of \( e^{-x} \))
- \( \frac{dv}{dx} = -\sin x \) (derivative of \( \cos x \))

Now applying the product rule:

\[ \frac{dy}{dx} = e^{-x}(-\sin x) + \cos x(-e^{-x}) \]

This simplifies to:

\[ \frac{dy}{dx} = -e^{-x} \sin x - e^{-x} \cos x \]

Factoring out \( -e^{-x} \):

\[ \frac{dy}{dx} = -e^{-x} (\sin x + \cos x) \]

### Step 2: Differentiate again to find the second derivative

Now we differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(-e^{-x} (\sin x + \cos x)\right) \]

Using the product rule again:

Let \( u = -e^{-x} \) and \( v = \sin x + \cos x \).

Now, we find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

- \( \frac{du}{dx} = e^{-x} \) (derivative of \( -e^{-x} \))
- \( \frac{dv}{dx} = \cos x - \sin x \) (derivative of \( \sin x + \cos x \))

Now applying the product rule:

\[ \frac{d^2y}{dx^2} = -e^{-x}(\cos x - \sin x) + (-e^{-x})(\sin x + \cos x) \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = -e^{-x}(\cos x - \sin x + \sin x + \cos x) \]

Combining like terms:

\[ \frac{d^2y}{dx^2} = -e^{-x}(2\sin x) \]

Thus, the second derivative is:

\[ \frac{d^2y}{dx^2} = -2e^{-x} \sin x \]

### Final Answer
The second derivative of the function \( y = e^{-x} \cos x \) is:

\[ \frac{d^2y}{dx^2} = -2e^{-x} \sin x \]

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