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Given that $y=2x^3 - \frac{6}{x^2}$, where $x \neq 0$, (a) find $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 1

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$Given-that-$y=2x^3---\frac{6}{x^2}$,-where-$x-\neq-0$,----(a)-find-$\frac{dy}{dx}$-Edexcel-A-Level Maths Pure-Question 6-2006-Paper 1.png$

Given that $y=2x^3 - \frac{6}{x^2}$, where $x \neq 0$, (a) find $\frac{dy}{dx}$. (b) find $\int y \: dx$.

Worked Solution & Example Answer:Given that $y=2x^3 - \frac{6}{x^2}$, where $x \neq 0$, (a) find $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 1

Step 1

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Answer

To find the derivative of $y = 2x^3 - \frac{6}{x^2}$ , we apply the power rule for differentiation.

Using the power rule for each term, we have:

The derivative of $2x^3$ is $6x^2$ .
The term $-\frac{6}{x^2}$ can be rewritten as $-6x^{-2}$ , and its derivative using the power rule is $12x^{-3}$ .

Combining these, we get:

$\frac{dy}{dx} = 6x^2 + 12x^{-3} = 6x^2 + \frac{12}{x^3}.$

Step 2

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Answer

To find the integral of $y = 2x^3 - \frac{6}{x^2}$ , we integrate each component separately:

For the term $2x^3$ , the integral is $\frac{2x^{4}}{4} = \frac{x^{4}}{2}$ .
The term $-\frac{6}{x^2}$ can be expressed as $-6x^{-2}$ , whose integral is $\frac{-6x^{-1}}{-1} = 6x^{-1} = \frac{6}{x}$ .

Thus, combining these results, we have:

$\int y \, dx = \frac{x^4}{2} + 6x^{-1} + C = \frac{x^4}{2} + \frac{6}{x} + C.$

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