If \(y = 2(2x + \sqrt{x})^{2}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
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Correct Answer: Option B
Explanation:
\(y = 2(2x + \sqrt{x})^{2}\)
Let \(u = 2x + \sqrt{x}\)
\(y = 2u^{2}\)
\(\frac{\mathrm d y}{\mathrm d u} = 4u\)
\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)
\(\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
= \(4u(2 + \frac{1}{2\sqrt{x}}) \)
= \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)
\(y = 2(2x + \sqrt{x})^{2}\)
Let \(u = 2x + \sqrt{x}\)
\(y = 2u^{2}\)
\(\frac{\mathrm d y}{\mathrm d u} = 4u\)
\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)
\(\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
= \(4u(2 + \frac{1}{2\sqrt{x}}) \)
= \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)