Find `dy/dx if 2x-3y=sinx-cosx`

Answer

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

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ 2x - 3y = \sin x - \cos x \]

2. **Differentiate both sides with respect to \(x\):**
- The left-hand side:
- The derivative of \(2x\) is \(2\).
- The derivative of \(-3y\) is \(-3\frac{dy}{dx}\) (using the chain rule).
- The right-hand side:
- The derivative of \(\sin x\) is \(\cos x\).
- The derivative of \(-\cos x\) is \(\sin x\).

Thus, differentiating gives us:
\[ 2 - 3\frac{dy}{dx} = \cos x + \sin x \]

3. **Rearranging the equation to isolate \(\frac{dy}{dx}\):**
\[ -3\frac{dy}{dx} = \cos x + \sin x - 2 \]
Now, multiply both sides by \(-1\):
\[ 3\frac{dy}{dx} = 2 - \cos x - \sin x \]

4. **Dividing both sides by \(3\):**
\[ \frac{dy}{dx} = \frac{2 - \cos x - \sin x}{3} \]

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