Given that y = 3x² + 6x + \frac{1}{x} + \frac{2x^3 - 7}{3\sqrt{x}}, x > 0 - Edexcel - A-Level Maths Pure - Question 9 - 2016 - Paper 1

To find the derivative ( \frac{dy}{dx} ), we will differentiate each term separately.

1. Differentiate ( 3x^2 ): [ \frac{d}{dx}(3x^2) = 6x ]

2. Differentiate ( 6x ): [ \frac{d}{dx}(6x) = 6 ]

3. Differentiate ( \frac{1}{x} ): Using the power rule, this can be written as ( x^{-1} ): [ \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2} ]

4. Differentiate ( \frac{2x^3 - 7}{3\sqrt{x}} ): This can be simplified to ( \frac{2x^3 - 7}{3x^{1/2}} ). Now, applying the quotient rule: Let ( u = 2x^3 - 7 ) and ( v = 3x^{1/2} ), then ( \frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2} ).

   • Calculate ( u' = 6x^2 )
   • Calculate ( v' = \frac{3}{2} x^{-1/2} = \frac{3}{2\sqrt{x}} )

   Therefore, applying the quotient rule, [ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{3x^{1/2}(6x^2) - (2x^3 - 7)(\frac{3}{2\sqrt{x}})}{(3x^{1/2})^2} ] Simplifying this:

   • Numerator: ( 18x^{5/2} - \frac{3(2x^3 - 7)}{2\sqrt{x}} = 18x^{5/2} - 3x^{5/2} + \frac{21}{2\sqrt{x}} ) = ( 15x^{5/2} + \frac{21}{2\sqrt{x}} )
   • Denominator: ( 9x ) Combining gives: [ \frac{15x^{5/2} + \frac{21}{2\sqrt{x}}}{9x} ] This further simplifies to:\n [ \frac{15x^{3/2}}{9} + \frac{7}{6x} ] \text{ (after simplification) }

Putting it all together: [ \frac{dy}{dx} = 6x + 6 - \frac{1}{x^2} + \frac{15x^{3/2}}{9} + \frac{7}{6x} ]