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: B
Explanation
\(y = 2(2x + \sqrt{x})^{2}\)
Let \(u = 2x + \sqrt{x}\)
\(y = 2u^{2}\)
\(\frac{\mathrm d y}{\mathrm d u} = 4u\)
\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)
\(\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
= \(4u(2 + \frac{1}{2\sqrt{x}}) \)
= \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)
