If `y=cos(log x)`, then `x^(2)(d^(2)y)/(dx^(2))+x(dy)/(dx)+y` is equal to

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C. 2
D. 3

Answer

To solve the problem, we need to find the expression \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y \) when \( y = \cos(\log x) \).

### Step 1: Differentiate \( y \) with respect to \( x \)

Given:
\[ y = \cos(\log x) \]

Using the chain rule to differentiate:
\[ \frac{dy}{dx} = -\sin(\log x) \cdot \frac{d}{dx}(\log x) \]

Since \( \frac{d}{dx}(\log x) = \frac{1}{x} \), we have:
\[ \frac{dy}{dx} = -\sin(\log x) \cdot \frac{1}{x} = -\frac{\sin(\log x)}{x} \]

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

Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{\sin(\log x)}{x}\right) \]

Using the quotient rule:
\[ \frac{d^2y}{dx^2} = -\left(\frac{x \cdot \frac{d}{dx}(\sin(\log x)) - \sin(\log x) \cdot \frac{d}{dx}(x)}{x^2}\right) \]

Calculating \( \frac{d}{dx}(\sin(\log x)) \):
\[ \frac{d}{dx}(\sin(\log x)) = \cos(\log x) \cdot \frac{1}{x} = \frac{\cos(\log x)}{x} \]

Substituting back:
\[ \frac{d^2y}{dx^2} = -\left(\frac{x \cdot \frac{\cos(\log x)}{x} - \sin(\log x) \cdot 1}{x^2}\right) = -\left(\frac{\cos(\log x) - \sin(\log x)}{x^2}\right) \]

Thus,
\[ \frac{d^2y}{dx^2} = -\frac{\cos(\log x) - \sin(\log x)}{x^2} \]

### Step 3: Substitute \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the expression

Now we substitute \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the expression:
\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y \]

Substituting:
\[ x^2 \left(-\frac{\cos(\log x) - \sin(\log x)}{x^2}\right) + x \left(-\frac{\sin(\log x)}{x}\right) + \cos(\log x) \]

This simplifies to:
\[ -(\cos(\log x) - \sin(\log x)) - \sin(\log x) + \cos(\log x) \]

Combining like terms:
\[ -\cos(\log x) + \sin(\log x) - \sin(\log x) + \cos(\log x) = 0 \]

### Final Result
Thus, we find that:
\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \]