if `f'(x) = sin ( log x) and y = f((2x +3)/(3-2x))` then ` (dy)/(dx)` equals

Options

A. `sin (log x) (1)/( x log x)`
B. ` (12)/((3-2x)^2) sin [ log ((2x +3)/(3-2x ))]`
C. `sin [ log ((2x +3)/(3-2x))]`
D. None of these

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given that \(f'(x) = \sin(\log x)\) and \(y = f\left(\frac{2x + 3}{3 - 2x}\right)\).

### Step-by-Step Solution:

1. **Identify the function and its derivative**:
We know that \(f'(x) = \sin(\log x)\).

2. **Apply the chain rule**:
To find \(\frac{dy}{dx}\), we will use the chain rule:
\[ \frac{dy}{dx} = f'\left(\frac{2x + 3}{3 - 2x}\right) \cdot \frac{d}{dx}\left(\frac{2x + 3}{3 - 2x}\right) \]

3. **Differentiate the inner function**:
Let \(u = \frac{2x + 3}{3 - 2x}\). We will differentiate \(u\) with respect to \(x\) using the quotient rule:
\[ \frac{du}{dx} = \frac{(3 - 2x)(2) - (2x + 3)(-2)}{(3 - 2x)^2} \]
Simplifying the numerator:
\[ = \frac{(6 - 4x + 4x + 6)}{(3 - 2x)^2} = \frac{12}{(3 - 2x)^2} \]

4. **Substituting back into the chain rule**:
Now substitute \(u\) back into the derivative:
\[ \frac{dy}{dx} = f'\left(\frac{2x + 3}{3 - 2x}\right) \cdot \frac{12}{(3 - 2x)^2} \]

5. **Substituting \(f'\)**:
Since \(f'(x) = \sin(\log x)\), we substitute:
\[ f'\left(\frac{2x + 3}{3 - 2x}\right) = \sin\left(\log\left(\frac{2x + 3}{3 - 2x}\right)\right) \]

6. **Final expression for \(\frac{dy}{dx}\)**:
Thus, we have:
\[ \frac{dy}{dx} = \sin\left(\log\left(\frac{2x + 3}{3 - 2x}\right)\right) \cdot \frac{12}{(3 - 2x)^2} \]

### Final Answer:
\[ \frac{dy}{dx} = \sin\left(\log\left(\frac{2x + 3}{3 - 2x}\right)\right) \cdot \frac{12}{(3 - 2x)^2} \]