For a function y = x cos x then ` ( d^(2) y)/( dx^(2))` is equal to

Options

A. ` - x cos x - 2 sin x `
B. ` x cos x + 2 sin x `
C. ` x sin x + cos x `
D. ` - x sin x + cos x `

Answer

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

### Step 1: Differentiate the function \( y = x \cos x \) to find the first derivative \( \frac{dy}{dx} \).

Using the product rule, which states that if \( u \) and \( v \) are functions of \( x \), then \( \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).

Let:
- \( u = x \) (so \( \frac{du}{dx} = 1 \))
- \( v = \cos x \) (so \( \frac{dv}{dx} = -\sin x \))

Now applying the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x(-\sin x) + \cos x(1) \]
This simplifies to:
\[ \frac{dy}{dx} = -x \sin x + \cos x \]

### Step 2: Differentiate \( \frac{dy}{dx} \) to find the second derivative \( \frac{d^2y}{dx^2} \).

Now we need to differentiate \( \frac{dy}{dx} = -x \sin x + \cos x \).

Again, we will use the product rule on the term \( -x \sin x \):
Let:
- \( u = -x \) (so \( \frac{du}{dx} = -1 \))
- \( v = \sin x \) (so \( \frac{dv}{dx} = \cos x \))

Applying the product rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-x \sin x) + \frac{d}{dx}(\cos x) \]
Calculating the first part:
\[ \frac{d}{dx}(-x \sin x) = -x \cos x + (-1)(\sin x) = -x \cos x - \sin x \]
And for the second part:
\[ \frac{d}{dx}(\cos x) = -\sin x \]

Combining these results gives:
\[ \frac{d^2y}{dx^2} = (-x \cos x - \sin x) + (-\sin x) = -x \cos x - 2\sin x \]

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

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