Find `dy/dx if ax+by^2=cosy`

Answer

To find \(\frac{dy}{dx}\) for the equation \(ax + by^2 = \cos y\), we will differentiate both sides with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate both sides of the equation:**
\[ \frac{d}{dx}(ax + by^2) = \frac{d}{dx}(\cos y) \]

2. **Differentiate the left-hand side:**
- The derivative of \(ax\) with respect to \(x\) is \(a\) (since \(a\) is a constant).
- For \(by^2\), we apply the chain rule:
\[ \frac{d}{dx}(by^2) = b \cdot 2y \cdot \frac{dy}{dx} = 2by \frac{dy}{dx} \]
- Therefore, the left-hand side becomes:
\[ a + 2by \frac{dy}{dx} \]

3. **Differentiate the right-hand side:**
- The derivative of \(\cos y\) is \(-\sin y\) multiplied by \(\frac{dy}{dx}\) (again using the chain rule):
\[ \frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx} \]

4. **Set the derivatives equal to each other:**
\[ a + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx} \]

5. **Rearrange the equation to isolate \(\frac{dy}{dx}\):**
- Move all terms involving \(\frac{dy}{dx}\) to one side:
\[ 2by \frac{dy}{dx} + \sin y \frac{dy}{dx} = -a \]

6. **Factor out \(\frac{dy}{dx}\):**
\[ \frac{dy}{dx}(2by + \sin y) = -a \]

7. **Solve for \(\frac{dy}{dx}\):**
\[ \frac{dy}{dx} = \frac{-a}{2by + \sin y} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{-a}{2by + \sin y} \]