If \(y = 2(2x + \sqrt{x})^{2}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
Answer Details
To differentiate the given function with respect to x, we will use the chain rule and the power rule of differentiation.
Let's start by using the power rule of differentiation to find the derivative of \((2x + \sqrt{x})^{2}\) with respect to x:
\[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\]
Next, we use the chain rule to differentiate \(\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\) with respect to x:
\[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x}) = 2 + \frac{1}{2\sqrt{x}}\]
Substituting this into the previous equation, we get:
\[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \left(2 + \frac{1}{2\sqrt{x}}\right)\]
Expanding this expression, we get:
\[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 4(2x + \sqrt{x}) + 2\sqrt{x}(2x + \sqrt{x})\]
Simplifying further, we get:
\[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 8x + 6\sqrt{x}\]
Finally, using the constant multiple rule of differentiation, we can find the derivative of the given function:
\[\frac{\mathrm d y}{\mathrm d x} = 2 \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(8x + 6\sqrt{x}) = 16x + 12\sqrt{x}\]
Therefore, the correct answer is: \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\).