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

Options

A. 0
B. 1
C. y
D. `-y`

Answer

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

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

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

Using the chain rule:
\[ \frac{dy}{dx} = -\sin(\log x) \cdot \frac{d}{dx}(\log x) \]
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: 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 product rule:
\[ \frac{d^2y}{dx^2} = -\left(\frac{d}{dx}(\sin(\log x)) \cdot \frac{1}{x} + \sin(\log x) \cdot \frac{d}{dx}\left(\frac{1}{x}\right)\right) \]

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

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

### Step 3: Substitute into the expression

Now we substitute \( \frac{d^2y}{dx^2} \) and \( \frac{dy}{dx} \) into the expression \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} \):
\[ x^2 \frac{d^2y}{dx^2} = x^2 \left(-\frac{\cos(\log x) - \sin(\log x)}{x^2}\right) = -(\cos(\log x) - \sin(\log x)) \]

And:
\[ x \frac{dy}{dx} = x \left(-\frac{\sin(\log x)}{x}\right) = -\sin(\log x) \]

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

### Step 4: Final expression

Since \( y = \cos(\log x) \), we can express the final result as:
\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = -y \]

### Conclusion

Thus, the final answer is:
\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = -y \]