WAEC - Further Mathematics (2024 - No. 31)
If y\(^2\) + 2xy - 8 = 0, find \(\frac{dy}{dx}\)
\(\frac{- y}{y + x}\)
\(\frac{2y + y}{y}\)
\(\frac{2y - x}{-y}\)
\(\frac{y}{2y - x}\)
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
\(\frac{dy}{dx}\) = \(\frac{-2y}{2(y + x)}\) = \(\frac{- y}{y + x}\)
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