Find `dy/dx if x+y^2=tanx+y`

Answer

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

### Step 1: Differentiate both sides of the equation with respect to \(x\).
Starting with the equation:
\[ x + y^2 = \tan x + y \]
We differentiate both sides:
\[ \frac{d}{dx}(x) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y) \]

### Step 2: Apply the differentiation rules.
The derivatives are:
- \(\frac{d}{dx}(x) = 1\)
- For \(y^2\), we use the chain rule: \(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\)
- \(\frac{d}{dx}(\tan x) = \sec^2 x\)
- For \(y\), we again use the chain rule: \(\frac{d}{dx}(y) = \frac{dy}{dx}\)

Putting it all together, we have:
\[ 1 + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \]

### Step 3: Rearrange the equation to isolate \(\frac{dy}{dx}\).
Now, we will collect all terms involving \(\frac{dy}{dx}\) on one side:
\[ 2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - 1 \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(2y - 1) = \sec^2 x - 1 \]

### Step 4: Solve for \(\frac{dy}{dx}\).
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\sec^2 x - 1}{2y - 1} \]

### Final Answer:
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\sec^2 x - 1}{2y - 1} \]

---