Given y = 3\sqrt{\text{x}} - 6\text{x} + 4, \quad x > 0 (a) find \int y \, dx, simplifying each term - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 1

To find \frac{dy}{dx}, we differentiate the given function for \text{y}:

$$\frac{dy}{dx} = \frac{d}{dx}(3\sqrt{x} - 6x + 4)$$

Calculating term by term, we have:

1. For ( 3\sqrt{x} ):

   $$\frac{d}{dx}(3\sqrt{x}) = 3 \cdot \frac{1}{2}x^{-1/2} = \frac{3}{2\sqrt{x}}$$

2. For ( -6x ):

   $$\frac{d}{dx}(-6x) = -6$$

3. For the constant ( 4 ):

   $$\frac{d}{dx}(4) = 0$$

Thus, combining the derivatives gives:

$$\frac{dy}{dx} = \frac{3}{2\sqrt{x}} - 6$$

To find the value of \text{x} for which \frac{dy}{dx} = 0, we set the expression equal to zero:

$$0 = \frac{3}{2\sqrt{x}} - 6$$

Solving for \text{x}, we first isolate \frac{3}{2\sqrt{x}}:

$$\frac{3}{2\sqrt{x}} = 6$$

Next, we multiply both sides by ( 2\sqrt{x} ) to eliminate the fraction:

$$3 = 12\sqrt{x}$$

Dividing both sides by 12 gives:

\sqrt{x} = \frac{1}{4}\n$$ Now squaring both sides yields:

\text{x} = \left(\frac{1}{4}\right)^2 = \frac{1}{16}

Thus, the value of \text{x} such that \frac{dy}{dx} = 0 is \( x = \frac{1}{16} \).