A-Level Edexcel Maths Pure Quadratics: Given that $y = 3x^2 + 4 oot{x},

Given that $y = 3x^2 + 4 oot{x}, \, x > 0$, find (a) $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 1

Question 6

$Given-that-$y-=-3x^2-+-4-oot{x},-\,-x->-0$,-find-(a)-$\frac{dy}{dx}$-Edexcel-A-Level Maths Pure-Question 6-2007-Paper 1.png$

Given that $y = 3x^2 + 4 oot{x}, \, x > 0$, find (a) $\frac{dy}{dx}$. (b) $\frac{d^2y}{dx^2}$. (c) $\int y \, dx$.

Worked Solution & Example Answer:Given that $y = 3x^2 + 4 oot{x}, \, x > 0$, find (a) $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 1

Step 1

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Answer

To find the derivative of $y$ with respect to $x$ , we will differentiate each term in the function:

The derivative of $3x^2$ is $6x$ .
For the term $4\root{x}$ , or equivalently $4x^{1/2}$ , we apply the power rule: $\frac{d}{dx}(4x^{1/2}) = 4 \cdot \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x}}.$

Combining these results, we obtain: $\frac{dy}{dx} = 6x + \frac{2}{\sqrt{x}}.$

Step 2

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Answer

To find the second derivative, we differentiate $\frac{dy}{dx}$ :

The derivative of $6x$ is simply $6$ .
For $\frac{2}{\sqrt{x}}$ , or $2x^{-1/2}$ , the derivative is: $\frac{d}{dx}(2x^{-1/2}) = 2 \cdot \left(-\frac{1}{2} \right)x^{-3/2} = -\frac{1}{\sqrt{x^3}}.$

Combining these results gives: $\frac{d^2y}{dx^2} = 6 - \frac{1}{\sqrt{x^3}}.$

Step 3

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Answer

To find the integral of $y$ , we integrate each term:

The integral of $3x^2$ is: $\int 3x^2 \, dx = x^3 + C.$
For $4\root{x}$ , or $4x^{1/2}$ : $\int 4x^{1/2} \, dx = \frac{4}{\frac{3}{2}}x^{\frac{3}{2}} = \frac{8}{3}x^{\frac{3}{2}} + C.$

Thus, the integral is: $\int y \, dx = x^3 + \frac{8}{3}x^{\frac{3}{2}} + C.$