If y ` =sin (x^(2) +x) ,then (dy)/(dx) =`

Options

A. ` -(2x +1) cos (x^(2) +x) `
B. ` (2x +1)cos (x^(2) +x) `
C. `2xcos (x^(2) +5) `
D. ` (2x+ 5)cos (x^(2) +5)`

Answer

To solve the problem where \( y' = \sin(x^2 + x) \), we need to find \( \frac{dy}{dx} \).

### Step-by-step Solution:

1. **Identify the function**:
We have \( y = \sin(x^2 + x) \).

2. **Differentiate using the chain rule**:
To differentiate \( y \), we will use the chain rule. The chain rule states that if you have a function \( y = \sin(u) \) where \( u = x^2 + x \), then:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

3. **Differentiate \( y \) with respect to \( u \)**:
The derivative of \( \sin(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = \cos(u) \]

4. **Differentiate \( u \) with respect to \( x \)**:
Now we need to find \( \frac{du}{dx} \) where \( u = x^2 + x \):
\[ \frac{du}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(x) = 2x + 1 \]

5. **Combine the derivatives**:
Now substitute \( u \) back into the equation:
\[ \frac{dy}{dx} = \cos(x^2 + x) \cdot (2x + 1) \]

6. **Final answer**:
Therefore, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = (2x + 1) \cos(x^2 + x) \]

### Summary:
The final result is:
\[ \frac{dy}{dx} = (2x + 1) \cos(x^2 + x) \]