If `y=sin(log_(e)x)`, then `x^(2)(d^(2)y)/(dx^(2))+x(dy)/(dx)` is equal to

Options

A. `sin(log_(e)x)`
B. `cos(log_(e)x)`
C. `y^(2)`
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 \( y = \sin(\log_e x) \).

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

Given:
\[ y = \sin(\log_e x) \]

Using the chain rule:
\[ \frac{dy}{dx} = \cos(\log_e x) \cdot \frac{d}{dx}(\log_e x) \]

We know:
\[ \frac{d}{dx}(\log_e x) = \frac{1}{x} \]

So,
\[ \frac{dy}{dx} = \cos(\log_e x) \cdot \frac{1}{x} = \frac{\cos(\log_e x)}{x} \]

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

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

Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{x \cdot \frac{d}{dx}(\cos(\log_e x)) - \cos(\log_e x) \cdot \frac{d}{dx}(x)}{x^2} \]

We need to compute \( \frac{d}{dx}(\cos(\log_e x)) \):
\[ \frac{d}{dx}(\cos(\log_e x)) = -\sin(\log_e x) \cdot \frac{d}{dx}(\log_e x) = -\sin(\log_e x) \cdot \frac{1}{x} \]

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

### 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 \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} \):
\[ x^2 \frac{d^2y}{dx^2} = x^2 \cdot \frac{-\sin(\log_e x) - \cos(\log_e x)}{x^2} = -\sin(\log_e x) - \cos(\log_e x) \]
\[ x \frac{dy}{dx} = x \cdot \frac{\cos(\log_e x)}{x} = \cos(\log_e x) \]

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

### Final Result
Thus, the final answer is:
\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = -\sin(\log_e x) \]

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