WAEC - Further Mathematics (2013 - No. 11)
Differentiate \(x^{2} + xy - 5 = 0\).
\(\frac{-(2x + y)}{x}\)
\(\frac{(2x - y)}{x}\)
\(\frac{-x}{2x + y}\)
\(\frac{(2x + y)}{x}\)
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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