Let `f'(x) = sin(x^2) and y = f(x^2 +1)` then `dy/dx` at `x=1` is

Options

A. `2 sin 2`
B. `2 cos 2`
C. `2 sin 4`
D. `cos 2`

Answer

To solve the problem step by step, we need to find \( \frac{dy}{dx} \) at \( x = 1 \) given that \( f'(x) = \sin(x^2) \) and \( y = f(x^2 + 1) \).

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

Given:
\[ y = f(x^2 + 1) \]

Using the chain rule, we differentiate \( y \):
\[ \frac{dy}{dx} = f'(x^2 + 1) \cdot \frac{d}{dx}(x^2 + 1) \]

### Step 2: Differentiate \( x^2 + 1 \)

Now, we differentiate \( x^2 + 1 \):
\[ \frac{d}{dx}(x^2 + 1) = 2x \]

### Step 3: Substitute back into the derivative

Substituting this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = f'(x^2 + 1) \cdot 2x \]

### Step 4: Find \( f'(x^2 + 1) \)

From the problem, we know:
\[ f'(x) = \sin(x^2) \]

Thus,
\[ f'(x^2 + 1) = \sin((x^2 + 1)^2) \]

### Step 5: Substitute \( f'(x^2 + 1) \) into \( \frac{dy}{dx} \)

Now we can write:
\[ \frac{dy}{dx} = 2x \cdot \sin((x^2 + 1)^2) \]

### Step 6: Evaluate \( \frac{dy}{dx} \) at \( x = 1 \)

Now we need to evaluate \( \frac{dy}{dx} \) at \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 2(1) \cdot \sin((1^2 + 1)^2) \]
\[ = 2 \cdot \sin((1 + 1)^2) \]
\[ = 2 \cdot \sin(2^2) \]
\[ = 2 \cdot \sin(4) \]

### Final Answer

Thus, the value of \( \frac{dy}{dx} \) at \( x = 1 \) is:
\[ \boxed{2 \sin(4)} \]