Find the derivatives of the following :
`x^(3)+sinx`

Answer

To find the derivative of the function \( y = x^3 + \sin x \), we will apply the rules of differentiation.

### Step-by-Step Solution:

1. **Identify the function**:
We have the function \( y = x^3 + \sin x \).

2. **Differentiate \( x^3 \)**:
Using the power rule of differentiation, which states that if \( y = x^n \), then \( \frac{dy}{dx} = n \cdot x^{n-1} \):
\[ \frac{d}{dx}(x^3) = 3 \cdot x^{3-1} = 3x^2 \]

3. **Differentiate \( \sin x \)**:
The derivative of \( \sin x \) is given by:
\[ \frac{d}{dx}(\sin x) = \cos x \]

4. **Combine the derivatives**:
Now, we can combine the results from steps 2 and 3:
\[ \frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(\sin x) = 3x^2 + \cos x \]

5. **Final result**:
Therefore, the derivative of the function \( y = x^3 + \sin x \) is:
\[ \frac{dy}{dx} = 3x^2 + \cos x \]