If \(y = 2(2x + \sqrt{x})^{2}\), find \(\frac{\mathrm d y}{\mathrm d x}\).

  • A \(2\sqrt{x}(2x + \sqrt{2})\)
  • B \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)
  • C \(4(2x + \sqrt{x})(2 + \sqrt{x})\)
  • D \(8(2x + \sqrt{x})(2 + \sqrt{x})\)
waec 2010furthermathematics

The correct answer is B. \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)

\(y = 2(2x + \sqrt{x})^{2}\)

Let \(u = 2x + \sqrt{x}\)

\(y = 2u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 4u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)

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

= \(4u(2 + \frac{1}{2\sqrt{x}}) \)

= \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)

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