Given that y = 3x^2 + 6x + \frac{1}{x} + \frac{2x^3 - 7}{3\sqrt{x}} , \quad x > 0 find \frac{dy}{dx} - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 1

To find ( \frac{dy}{dx} ), we differentiate each term in the function separately:

1. ( 3x^2 ) becomes ( 6x ).

2. ( 6x ) becomes ( 6 ).

3. ( \frac{1}{x} ) or ( x^{-1} ) becomes ( -x^{-2} ) (or ( -\frac{1}{x^2} )).

4. For the term ( \frac{2x^3 - 7}{3\sqrt{x}} ), we apply the Quotient Rule:

   • Let ( u = 2x^3 - 7 ) and ( v = 3\sqrt{x} ).
   • Then, ( \frac{d}{dx}(u/v) = \frac{u'v - uv'}{v^2} ).

   Calculating each derivative:

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

   Applying the Quotient Rule:

   • ( \frac{d}{dx}(\frac{2x^3 - 7}{3\sqrt{x}}) = \frac{(6x^2)(3\sqrt{x}) - (2x^3 - 7)(\frac{3}{2\sqrt{x}})}{(3\sqrt{x})^2} )

   Simplifying this gives:

   • The numerator expands to ( 18x^{5/2} - \frac{3(2x^3 - 7)}{2} = 18x^{5/2} - 3x^3 + \frac{21}{2\sqrt{x}} )
   • The denominator is ( 9x )

Thus, the term is ( \frac{18x^{5/2} - 3x^3 + \frac{21}{2\sqrt{x}}}{9x} ).