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.

\(8(2x + \sqrt{x})(2 + \sqrt{x})\)

Correct answer is B

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

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

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

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

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

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

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

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