find `dy/dx`
`ysecx-y^2cosx+2x=0`

Answer

To find \(\frac{dy}{dx}\) for the equation \(y \sec x - y^2 \cos x + 2x = 0\), we will differentiate both sides of the equation with respect to \(x\) using implicit differentiation. Here are the steps:

### Step 1: Differentiate the equation
We start with the equation:
\[ y \sec x - y^2 \cos x + 2x = 0 \]
Differentiating each term with respect to \(x\):
- The derivative of \(y \sec x\) using the product rule is:
\[ \frac{d}{dx}(y \sec x) = y \frac{d}{dx}(\sec x) + \sec x \frac{dy}{dx} = y \sec x \tan x + \sec x \frac{dy}{dx} \]
- The derivative of \(-y^2 \cos x\) using the product rule is:
\[ \frac{d}{dx}(-y^2 \cos x) = -\left(2y \frac{dy}{dx} \cos x + y^2 \frac{d}{dx}(\cos x)\right) = -\left(2y \frac{dy}{dx} \cos x - y^2 \sin x\right) = -2y \frac{dy}{dx} \cos x + y^2 \sin x \]
- The derivative of \(2x\) is:
\[ \frac{d}{dx}(2x) = 2 \]
- The derivative of \(0\) is \(0\).

Putting it all together, we have:
\[ y \sec x \tan x + \sec x \frac{dy}{dx} - 2y \frac{dy}{dx} \cos x + y^2 \sin x + 2 = 0 \]

### Step 2: Combine like terms
Now, we can group the terms involving \(\frac{dy}{dx}\):
\[ \sec x \frac{dy}{dx} - 2y \cos x \frac{dy}{dx} = -y \sec x \tan x - y^2 \sin x - 2 \]
Factoring out \(\frac{dy}{dx}\) from the left side:
\[ \frac{dy}{dx}(\sec x - 2y \cos x) = -y \sec x \tan x - y^2 \sin x - 2 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Now we can isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{-y \sec x \tan x - y^2 \sin x - 2}{\sec x - 2y \cos x} \]

### Step 4: Simplify the expression
To make it cleaner, we can factor out a negative sign from the numerator:
\[ \frac{dy}{dx} = \frac{y \sec x \tan x + y^2 \sin x + 2}{2y \cos x - \sec x} \]

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

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