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

To find ( \frac{dy}{dx} ), we will differentiate each term in the expression for ( y ):

1. For the term ( 3x^2 ), the derivative is: $\frac{d}{dx}(3x^2) = 6x$

2. For the term ( 6x^{\frac{1}{3}} ), we apply the power rule: $\frac{d}{dx}(6x^{\frac{1}{3}}) = 6 \cdot \frac{1}{3} x^{-\frac{2}{3}} = 2x^{-\frac{2}{3}}$

3. For the term ( \frac{2x^3 - 7}{3\text{√}x} ), we will use the Quotient Rule. Let ( u = 2x^3 - 7 ) and ( v = 3x^{\frac{1}{2}} ):

   • First, find ( \frac{du}{dx} = 6x^2 )
   • Next, find ( \frac{dv}{dx} = \frac{3}{2} x^{-\frac{1}{2}} )
   • Apply the Quotient Rule: $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = \frac{3x^{\frac{1}{2}} (6x^2) - (2x^3 - 7)(\frac{3}{2} x^{-\frac{1}{2}})}{(3x^{\frac{1}{2}})^2}$