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

Answer

To find \(\frac{dy}{dx}\) for the equation \(2x - 3y = \tan x\), we will use implicit differentiation. Here are the steps:

### Step 1: Differentiate both sides of the equation
We start with the equation:
\[ 2x - 3y = \tan x \]
Now, we differentiate both sides with respect to \(x\).

### Step 2: Differentiate the left side
The left side is \(2x - 3y\). We differentiate each term:
- The derivative of \(2x\) is \(2\).
- The derivative of \(-3y\) is \(-3\frac{dy}{dx}\) (using the chain rule since \(y\) is a function of \(x\)).

So, the differentiation of the left side gives us:
\[ \frac{d}{dx}(2x - 3y) = 2 - 3\frac{dy}{dx} \]

### Step 3: Differentiate the right side
The right side is \(\tan x\). The derivative of \(\tan x\) is \(\sec^2 x\).

### Step 4: Set the derivatives equal
Now, we equate the derivatives from both sides:
\[ 2 - 3\frac{dy}{dx} = \sec^2 x \]

### Step 5: Solve for \(\frac{dy}{dx}\)
We need to isolate \(\frac{dy}{dx}\):
1. Rearranging gives:
\[ -3\frac{dy}{dx} = \sec^2 x - 2 \]
2. Now, divide both sides by \(-3\):
\[ \frac{dy}{dx} = \frac{2 - \sec^2 x}{3} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{2 - \sec^2 x}{3} \]