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Given that $y=2x^6 + 7 + \frac{1}{x^3}, \ x \neq 0$, find, in their simplest form, (a) $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 1

Question 4

$Given-that-$y=2x^6-+-7-+-\frac{1}{x^3},-\-x-\neq-0$,-find,-in-their-simplest-form,---(a)-$\frac{dy}{dx}$-Edexcel-A-Level Maths Pure-Question 4-2011-Paper 1.png$

Given that $y=2x^6 + 7 + \frac{1}{x^3}, \ x \neq 0$, find, in their simplest form, (a) $\frac{dy}{dx}$. (b) $\int y \, dx$.

Worked Solution & Example Answer:Given that $y=2x^6 + 7 + \frac{1}{x^3}, \ x \neq 0$, find, in their simplest form, (a) $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 1

Step 1

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

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Answer

To find $\frac{dy}{dx}$ , we will differentiate each term in the equation:

The derivative of $2x^6$ is $12x^5$ .
The derivative of the constant $7$ is $0$ .
Applying the power rule to $\frac{1}{x^3}$ , we rewrite it as $x^{-3}$ and differentiate to get $-3x^{-4}$ .

Combining these terms, we have:

$\frac{dy}{dx} = 12x^5 - 3x^{-4} = 12x^5 - \frac{3}{x^4}$

Step 2

(b) $\int y \, dx$

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Answer

To integrate $y$ , we will integrate each term in the expression:

For $2x^6$ , the integral is: $\int 2x^6 \, dx = \frac{2}{7}x^7$
The integral of the constant $7$ is: $7x$
For $\frac{1}{x^3}$ or $x^{-3}$ , the integral is: $\int x^{-3} \, dx = -\frac{1}{2}x^{-2}$

Putting it all together, we get:

$\int y \, dx = \frac{2}{7}x^7 + 7x - \frac{1}{2}x^{-2} + C$

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