If ` Y=f ((2x -1)/(x^2 +1)) and f'(x) = sin x^2 ,` then ` (dy)/(dx)` at ` x=0` equals

Options

A. `1/2 sin 1 `
B. ` sin 1`
C. `2 sin 1`
D. None of these

Answer

To solve the problem step by step, we will use the chain rule and the quotient rule for differentiation.

### Step 1: Define the function and its derivative
Given:
\[ Y = f\left(\frac{2x - 1}{x^2 + 1}\right) \]
and
\[ f'(x) = \sin(x^2) \]

### Step 2: Differentiate using the chain rule
To find \(\frac{dy}{dx}\), we apply the chain rule:
\[ \frac{dy}{dx} = f'\left(\frac{2x - 1}{x^2 + 1}\right) \cdot \frac{d}{dx}\left(\frac{2x - 1}{x^2 + 1}\right) \]

### Step 3: Differentiate the inner function using the quotient rule
Let:
\[ u = 2x - 1 \]
\[ v = x^2 + 1 \]

Using the quotient rule:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \(\frac{du}{dx}\) and \(\frac{dv}{dx}\):
\[ \frac{du}{dx} = 2 \]
\[ \frac{dv}{dx} = 2x \]

Now substituting back:
\[ \frac{d}{dx}\left(\frac{2x - 1}{x^2 + 1}\right) = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} \]

### Step 4: Simplify the derivative
Expanding the numerator:
\[ = \frac{2(x^2 + 1) - (4x^2 - 2x)}{(x^2 + 1)^2} \]
\[ = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} \]
\[ = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \]

### Step 5: Substitute back into the derivative
Now substituting back into the expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = f'\left(\frac{2x - 1}{x^2 + 1}\right) \cdot \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \]

### Step 6: Evaluate at \(x = 0\)
First, calculate \(\frac{2(0) - 1}{0^2 + 1}\):
\[ = \frac{-1}{1} = -1 \]

Now, substitute \(x = 0\) into \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = f'(-1) \cdot \frac{-2(0)^2 + 2(0) + 2}{(0^2 + 1)^2} \]
\[ = f'(-1) \cdot \frac{2}{1^2} \]
\[ = f'(-1) \cdot 2 \]

Now, find \(f'(-1)\):
\[ f'(-1) = \sin((-1)^2) = \sin(1) \]

### Final Result
Thus:
\[ \frac{dy}{dx} \bigg|_{x=0} = 2 \sin(1) \]