Given that y = 3x^2 + 6x^{rac{1}{2}} + rac{2x^3 - 7}{3 ext{√}x}, x > 0 find \( \frac{dy}{dx} \) - Edexcel - A-Level Maths Pure - Question 7 - 2016 - Paper 1

We differentiate each term individually:

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

2. For the term ( 6x^{\frac{1}{2}} ): [ \frac{d}{dx}(6x^{\frac{1}{2}}) = 3x^{-\frac{1}{2}} ]

3. For the term ( \frac{2x^3 - 7}{3\text{√}x} ): We will use the quotient rule ( \frac{u}{v} ), where ( u = 2x^3 - 7 ) and ( v = 3\text{√}x = 3x^{\frac{1}{2}} ). The quotient rule states: [ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]

   • First, we need ( \frac{du}{dx} ): [ \frac{d}{dx}(2x^3 - 7) = 6x^2 ]
   • Next, we find ( \frac{dv}{dx} ): [ \frac{d}{dx}(3x^{\frac{1}{2}}) = \frac{3}{2} x^{-\frac{1}{2}} ]
   • Now, we can apply the quotient rule: [ \frac{d}{dx}\left( \frac{2x^3 - 7}{3x^{\frac{1}{2}}} \right) = \frac{3x^{\frac{1}{2}}(6x^2) - (2x^3 - 7)(\frac{3}{2} x^{-\frac{1}{2}})}{(3x^{\frac{1}{2}})^2} ]
   • Simplifying this: [ \frac{18x^{\frac{5}{2}} - \frac{3}{2}(2x^3 - 7)x^{-\frac{1}{2}}}{9x} = \frac{18x^{\frac{5}{2}} - \frac{3(2x^3 - 7)}{2x^{\frac{1}{2}}}}{9x} ]
   • This simplifies to: [ \frac{36x^{3} - (2x^3 - 7)}{18x^{\frac{3}{2}}} ]
   • Finally, aligning all terms we aggregate:
     • ( 6x + 3x^{-\frac{1}{2}} + \frac{36x^3 - (2x^3 - 7)}{18x^{\frac{3}{2}}} )

Therefore: [ \frac{dy}{dx} = 6x + 3x^{-\frac{1}{2}} + \frac{34x^3 + 7}{18x^{\frac{3}{2}}} ]