WAEC - Further Mathematics (2011 - No. 12)

\(f(x) = (x^{2} + 3)^{2}\)

Using the chain rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{\mathrm d y}{\mathrm d u} \times \frac{\mathrm d u}{\mathrm d x}\)

Let \(u = x^{2} + 3\) so that \(y = u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 2u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2x\)

\(\therefore \frac{\mathrm d y}{\mathrm d x} = 2u(2x) = 4xu\)

But \(u = x^{2} + 3\),

\(\frac{\mathrm d y}{\mathrm d x} = 4x(x^{2} + 3)\)

At \(x = \frac{1}{2}, \frac{\mathrm d y}{\mathrm d x} = 4(\frac{1}{2})((\frac{1}{2})^{2} + 3)\)

= \(2 \times \frac{13}{4} = \frac{13}{2} = 6.5\)