A-Level Edexcel Maths Pure Quadratics: Given that $y = 4x^3

Given that $y = 4x^3 - \frac{5}{x^2}, x \neq 0$, find in their simplest form (a) $ \frac{dy}{dx} $ (b) $ \int y \, dx $ - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 1

Question 5

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Given that $y = 4x^3 - \frac{5}{x^2}, x \neq 0$, find in their simplest form (a) $ \frac{dy}{dx} $ (b) $ \int y \, dx $

Worked Solution & Example Answer:Given that $y = 4x^3 - \frac{5}{x^2}, x \neq 0$, find in their simplest form (a) $ \frac{dy}{dx} $ (b) $ \int y \, dx $ - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 1

Step 1

(a) $ \frac{dy}{dx} $

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Answer

To find ( \frac{dy}{dx} ), we will use the power rule for differentiation. The given function is:

y = 4x^3 - \frac{5}{x^2} = 4x^3 - 5x^{-2}

Now, we can differentiate term by term:

The derivative of ( 4x^3 ) is ( 12x^2 ).
The derivative of ( -5x^{-2} ) is ( 10x^{-3} ) (using the power rule).

Combining these results:

\frac{dy}{dx} = 12x^2 + 10x^{-3}

We can simplify this further:

\frac{dy}{dx} = 12x^2 + \frac{10}{x^3}

Step 2

(b) $ \int y \, dx $

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Answer

To find the integral ( \int y , dx ), we use the function we have:

y = 4x^3 - \frac{5}{x^2}

Integrating term by term:

The integral of ( 4x^3 ) is ( \frac{4}{4}x^4 = x^4 ).
The integral of ( -\frac{5}{x^2} ) is ( -5 \int x^{-2} , dx = -5[-x^{-1}] = \frac{5}{x} ).

Putting this together:

\int y \, dx = x^4 + \frac{5}{x} + C

where ( C ) is the constant of integration.

Given that $y = 4x^3 - \frac{5}{x^2}, x \neq 0$, find in their simplest form (a) \( \frac{dy}{dx} \) (b) \( \int y \, dx \) - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 1

Worked Solution & Example Answer:Given that $y = 4x^3 - \frac{5}{x^2}, x \neq 0$, find in their simplest form (a) \( \frac{dy}{dx} \) (b) \( \int y \, dx \) - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 1

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

(b) \( \int y \, dx \)

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