Find `dy/dx if x=4y^2-sinx`

Answer

To find \( \frac{dy}{dx} \) given the equation \( x = 4y^2 - \sin x \), we will differentiate both sides of the equation with respect to \( x \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ x = 4y^2 - \sin x \]

2. **Differentiate both sides with respect to \( x \):**
- The derivative of \( x \) with respect to \( x \) is \( 1 \).
- For the term \( 4y^2 \), we apply the chain rule. The derivative is:
\[ \frac{d}{dx}(4y^2) = 4 \cdot 2y \cdot \frac{dy}{dx} = 8y \frac{dy}{dx} \]
- The derivative of \( -\sin x \) is \( -\cos x \).

Putting it all together, we have:
\[ 1 = 8y \frac{dy}{dx} - \cos x \]

3. **Rearranging the equation to solve for \( \frac{dy}{dx} \):**
\[ 8y \frac{dy}{dx} = 1 + \cos x \]
\[ \frac{dy}{dx} = \frac{1 + \cos x}{8y} \]

### Final Result:
Thus, the value of \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1 + \cos x}{8y} \]