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2014 WAEC Further Mathematics Theory Find the gradient of \(xy^{2} + x^{2} y = 4xy\) at the point (1, 3).

Further Mathematics
WAEC 2014

Find the gradient of \(xy^{2} + x^{2} y = 4xy\) at the point (1, 3).

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Explanation

\(xy^{2} + x^{2} y = 4xy\)

Differentiating with respect to x,

\(2xy \frac{\mathrm d y}{\mathrm d x} + y^{2} + x^{2} \frac{\mathrm d y}{\mathrm d x} + 2xy = 4x \frac{\mathrm d y}{\mathrm d x} + 4y\)

\((2xy + x^{2} - 4x) \frac{\mathrm d y}{\mathrm d x} = 4y - y^{2} - 2xy\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{4y^{2} - y^{2} - 2xy}{2xy + x^{2} - 4x}\)

At (1, 3), Gradient = \(\frac{4(3) - (3)^{2} - 2(1)(3)}{2(1)(3) + 1^{2} - 4(1)}\)

= \(\frac{-3}{3}\)

= -1


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