Find `dy/dx" when "2x+3y=sinx`.

Answer

To find \(\frac{dy}{dx}\) when \(2x + 3y = \sin x\), we will differentiate both sides of the equation with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate both sides of the equation**:
The given equation is:
\[ 2x + 3y = \sin x \]
We will differentiate each term with respect to \(x\).

2. **Differentiate \(2x\)**:
The derivative of \(2x\) with respect to \(x\) is:
\[ \frac{d}{dx}(2x) = 2 \]

3. **Differentiate \(3y\)**:
Since \(y\) is a function of \(x\), we use implicit differentiation:
\[ \frac{d}{dx}(3y) = 3\frac{dy}{dx} \]

4. **Differentiate \(\sin x\)**:
The derivative of \(\sin x\) with respect to \(x\) is:
\[ \frac{d}{dx}(\sin x) = \cos x \]

5. **Combine the derivatives**:
Putting it all together, we have:
\[ 2 + 3\frac{dy}{dx} = \cos x \]

6. **Isolate \(\frac{dy}{dx}\)**:
To find \(\frac{dy}{dx}\), we rearrange the equation:
\[ 3\frac{dy}{dx} = \cos x - 2 \]

7. **Divide by 3**:
Finally, we divide both sides by 3:
\[ \frac{dy}{dx} = \frac{\cos x - 2}{3} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{\cos x - 2}{3} \]