If ` f' (x) = sin (log x)` and `y = f((2x+3)/(3- 2x)), " then " (dy)/(dx) " at " x = 1 ` is equal to

Options

A. `cos(log5)`
B. `sin(log5)`
C. `12sin(log5)`
D. `5/12 sin(log5)`

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) at \(x = 1\) given that \(f'(x) = \sin(\log x)\) and \(y = f\left(\frac{2x+3}{3-2x}\right)\).

### Step 1: Differentiate \(y\) using the chain rule
We start by applying the chain rule to differentiate \(y\):
\[ \frac{dy}{dx} = f'\left(\frac{2x+3}{3-2x}\right) \cdot \frac{d}{dx}\left(\frac{2x+3}{3-2x}\right) \]

### Step 2: Find \(\frac{d}{dx}\left(\frac{2x+3}{3-2x}\right)\)
To differentiate \(\frac{2x+3}{3-2x}\), we use the quotient rule:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
where \(u = 2x + 3\) and \(v = 3 - 2x\).

Calculating \(u'\) and \(v'\):
- \(u' = 2\)
- \(v' = -2\)

Now applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{2x+3}{3-2x}\right) = \frac{(2)(3-2x) - (2x+3)(-2)}{(3-2x)^2} \]
\[ = \frac{6 - 4x + 4x + 6}{(3-2x)^2} = \frac{12}{(3-2x)^2} \]

### Step 3: Substitute back into \(\frac{dy}{dx}\)
Now substituting back into the derivative:
\[ \frac{dy}{dx} = f'\left(\frac{2x+3}{3-2x}\right) \cdot \frac{12}{(3-2x)^2} \]

### Step 4: Evaluate \(f'\left(\frac{2x+3}{3-2x}\right)\)
Since \(f'(x) = \sin(\log x)\), we need to evaluate:
\[ f'\left(\frac{2x+3}{3-2x}\right) = \sin\left(\log\left(\frac{2x+3}{3-2x}\right)\right) \]

### Step 5: Substitute \(x = 1\)
Now we substitute \(x = 1\):
\[ \frac{2(1)+3}{3-2(1)} = \frac{2 + 3}{3 - 2} = \frac{5}{1} = 5 \]
Thus,
\[ f'\left(5\right) = \sin(\log(5)) \]

### Step 6: Calculate \(\frac{dy}{dx}\) at \(x = 1\)
Now substituting \(x = 1\) into \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \sin(\log(5)) \cdot \frac{12}{(3-2(1))^2} = \sin(\log(5)) \cdot \frac{12}{1^2} = 12 \sin(\log(5)) \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(x = 1\) is:
\[ \frac{dy}{dx} = 12 \sin(\log(5)) \]