If y\(^2\) + 2xy – 8 = 0, find \(\frac{dy}{dx}\)
-
\(\frac{- y}{y + x}\)
- \(\frac{2y + y}{y}\)
- \(\frac{2y - x}{-y}\)
- \(\frac{y}{2y - x}\)
The correct answer is: A
Explanation
y\(^2\) + 2xy - 8 = 0
\(\frac{dy}{dx}\) = 2y\(\frac{dy}{dx}\) + 2y + 2x\(\frac{dy}{dx}\) = 0
= 2y\(\frac{dy}{dx}\) + 2x\(\frac{dy}{dx}\) = -2y
\(\frac{dy}{dx}\) = \(\frac{-2y}{2(y + x)}\) = \(\frac{- y}{y + x}\)