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Given that $y = 2x^4 - \frac{6}{x^2}$, where $x \neq 0$ - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 1

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Question 6

Given-that-$y-=-2x^4---\frac{6}{x^2}$,-where-$x-\neq-0$-Edexcel-A-Level Maths Pure-Question 6-2006-Paper 1.png

Given that $y = 2x^4 - \frac{6}{x^2}$, where $x \neq 0$. (a) find \( \frac{dy}{dx} \). (b) find \( \int y \; dx \).

Worked Solution & Example Answer:Given that $y = 2x^4 - \frac{6}{x^2}$, where $x \neq 0$ - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 1

Step 1

find \( \frac{dy}{dx} \)

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Answer

To find the derivative of ( y ) with respect to ( x ), we will apply the power rule of differentiation:

  1. Differentiate the first term: ( 2x^4 ) gives ( \frac{d}{dx}(2x^4) = 8x^3 ).
  2. Differentiate the second term: ( -\frac{6}{x^2} ) can be rewritten as ( -6x^{-2} ), leading to ( \frac{d}{dx}(-6x^{-2}) = 12x^{-3}. )

Thus, combining these gives:

dydx=8x3+12x3ext,or8x3+12x3.\frac{dy}{dx} = 8x^3 + 12x^{-3} ext{, or } 8x^3 + \frac{12}{x^3}.

Step 2

find \( \int y \; dx \)

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Answer

To find the integral of ( y ) with respect to ( x ), we will integrate each term:

  1. The integral of ( 2x^4 ) is ( \int 2x^4 , dx = \frac{2}{5}x^5 + C_1 ).
  2. The integral of ( -\frac{6}{x^2} ) is ( \int -6x^{-2} , dx = 6x^{-1} + C_2. )

Putting these together, we have:

ydx=25x5+6x1+C,\int y \, dx = \frac{2}{5}x^5 + 6x^{-1} + C,

where ( C ) is the constant of integration.

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