Find the second derivatives of the following functions :
`xcosx`

Answer

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

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

Given:
\[ y = x \cos x \]

Using the product rule for differentiation, where if \( u = x \) and \( v = \cos x \), then:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = -\sin x \)

Now applying the product rule:
\[ \frac{dy}{dx} = x(-\sin x) + \cos x(1) \]
\[ \frac{dy}{dx} = -x \sin x + \cos x \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -x \sin x + \cos x \]

Using the product rule again on the term \( -x \sin x \):
Let \( u = -x \) and \( v = \sin x \):
- \( \frac{du}{dx} = -1 \)
- \( \frac{dv}{dx} = \cos x \)

Applying the product rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-x \sin x) + \frac{d}{dx}(\cos x) \]
\[ \frac{d^2y}{dx^2} = (-x \cos x + \sin x) - \sin x \]
\[ \frac{d^2y}{dx^2} = -x \cos x \]

### Final Result
Thus, the second derivative of \( y = x \cos x \) is:
\[ \frac{d^2y}{dx^2} = -x \cos x \]