If `y sec x+ tan x +x^2 y=0`, then `(dy)/(dx)` =

Options

A. `(2xy+sec^2x+ysecxtanx)/(x^2+secx)`
B. `-(2xy+sec^2x+secxtanx)/(x^2+secx)`
C. `-(2xy+sec^2x+ysecxtanx)/(x^2+secx)`
D. None of these

Answer

To find \(\frac{dy}{dx}\) from the equation \(y \sec x + \tan x + x^2 y = 0\), we will use implicit differentiation. Here’s a step-by-step solution:

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

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

For \(\tan x\), the derivative is:
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]

For \(x^2 y\), we again apply the product rule:
\[ \frac{d}{dx}(x^2 y) = x^2 \frac{dy}{dx} + 2xy \]

### Step 3: Combine the derivatives
Now, substituting back into our differentiated equation, we get:
\[ y \sec x \tan x + \sec x \frac{dy}{dx} + \sec^2 x + x^2 \frac{dy}{dx} + 2xy = 0 \]

### Step 4: Collect all \(\frac{dy}{dx}\) terms
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ \sec x \frac{dy}{dx} + x^2 \frac{dy}{dx} = - (y \sec x \tan x + \sec^2 x + 2xy) \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} (\sec x + x^2) = - (y \sec x \tan x + \sec^2 x + 2xy) \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Finally, we solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = - \frac{y \sec x \tan x + \sec^2 x + 2xy}{\sec x + x^2} \]

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