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Differentiate \(x^{2} + xy – 5 = 0\).

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Further Mathematics WAEC 2013

Differentiate \(x^{2} + xy – 5 = 0\).

  • \(\frac{-(2x + y)}{x}\) checkmark
  • \(\frac{(2x - y)}{x}\)
  • \(\frac{-x}{2x + y}\)
  • \(\frac{(2x + y)}{x}\)

The correct answer is: A

Explanation

\(\frac{\mathrm d}{\mathrm d x}(x^2 + xy - 5) = \frac{\mathrm d (x^{2})}{\mathrm d x} + \frac{\mathrm d (xy)}{\mathrm d x} - \frac{\mathrm d (5)}{\mathrm d x} = 0\)

= \(2x + x\frac{\mathrm d y}{\mathrm d x} + y = 0\)

\(\implies x\frac{\mathrm d y}{\mathrm d x} = -(2x + y)\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{-(2x + y)}{x}\)

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