`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 with respect to \(x\). Here’s the step-by-step solution:

### Step 1: Differentiate both sides
We start with the equation:
\[ y \sec x - y^2 \cos x + 2x = 0 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(y \sec x) - \frac{d}{dx}(y^2 \cos x) + \frac{d}{dx}(2x) = 0 \]

### Step 2: Apply the product rule
For the first term \(y \sec x\), we apply the product rule:
\[ \frac{d}{dx}(y \sec x) = \frac{dy}{dx} \sec x + y \frac{d}{dx}(\sec x) = \frac{dy}{dx} \sec x + y \sec x \tan x \]

For the second term \(y^2 \cos x\), we also apply the product rule:
\[ \frac{d}{dx}(y^2 \cos x) = \frac{d}{dx}(y^2) \cos x + y^2 \frac{d}{dx}(\cos x) = 2y \frac{dy}{dx} \cos x - y^2 \sin x \]

The derivative of \(2x\) is simply:
\[ \frac{d}{dx}(2x) = 2 \]

### Step 3: Substitute the derivatives back into the equation
Substituting these derivatives back into our differentiated equation gives:
\[ \left(\frac{dy}{dx} \sec x + y \sec x \tan x\right) - \left(2y \frac{dy}{dx} \cos x - y^2 \sin x\right) + 2 = 0 \]

### Step 4: Rearranging the equation
Rearranging the equation, we have:
\[ \frac{dy}{dx} \sec x + y \sec x \tan x - 2y \frac{dy}{dx} \cos x + y^2 \sin x + 2 = 0 \]

### Step 5: Collecting \(\frac{dy}{dx}\) terms
Now, we will collect all terms involving \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \sec x - 2y \frac{dy}{dx} \cos x = -y \sec x \tan x - y^2 \sin x - 2 \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left(\sec x - 2y \cos x\right) = -\left(y \sec x \tan x + y^2 \sin x + 2\right) \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Finally, we can isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{y \sec x \tan x + y^2 \sin x + 2}{\sec x - 2y \cos x} \]

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