Given that $y = 2x^5 + \frac{6}{\sqrt{x}}$, $x > 0$, find in their simplest form (a) $\frac{dy}{dx}$ (b) $\int y \, dx$ - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 2

To integrate $y = 2x^5 + \frac{6}{\sqrt{x}}$:

1. Rewrite $\frac{6}{\sqrt{x}}$ as $6x^{-\frac{1}{2}}$.

2. Integrate the first term:

   • The integral of $2x^5$ is $\frac{2}{6}x^6 = \frac{1}{3}x^6$.

3. Integrate the second term:

   • The integral of $6x^{-\frac{1}{2}}$ is $6 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 12x^{\frac{1}{2}}$.

4. Combine the results and include the constant of integration $c$:

   $\int y \, dx = \frac{1}{3}x^6 + 12x^{\frac{1}{2}} + c$