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Given that $y = 2x^3 + \frac{6}{\sqrt{x}}$, $x > 0$, find in their simplest form - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 2
Question 6
Given that $y = 2x^3 + \frac{6}{\sqrt{x}}$, $x > 0$, find in their simplest form. (a) \(\frac{dy}{dx}\) (b) \(\int y \, dx\)
Worked Solution & Example Answer:Given that $y = 2x^3 + \frac{6}{\sqrt{x}}$, $x > 0$, find in their simplest form - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 2
Step 1
a) \(\frac{dy}{dx}\)
Answer
To find (\frac{dy}{dx}), we need to differentiate (y = 2x^3 + 6x^{-\frac{1}{2}}) with respect to (x).
- Differentiate (2x^3): (\frac{d}{dx}(2x^3) = 6x^2).
- Differentiate (6x^{-\frac{1}{2}}): (\frac{d}{dx}(6x^{-\frac{1}{2}}) = -3x^{-\frac{3}{2}}).
- Combine the results:
[ \frac{dy}{dx} = 6x^2 - 3x^{-\frac{3}{2}}. ]
Step 2
b) \(\int y \, dx\)
Answer
To integrate (y = 2x^3 + 6x^{-\frac{1}{2}}), we need to integrate each term separately:
-
Integrate (2x^3):
[ \int 2x^3 , dx = \frac{2}{4}x^4 = \frac{1}{2}x^4. ] -
Integrate (6x^{-\frac{1}{2}}):
[ \int 6x^{-\frac{1}{2}} , dx = 6 \cdot \left( 2x^{\frac{1}{2}} \right) = 12x^{\frac{1}{2}}. ] -
Combine results of integration and add the constant of integration (c):
[ \int y , dx = \frac{1}{2}x^4 + 12x^{\frac{1}{2}} + c. ]