A-Level Edexcel Maths Pure Transformations of Functions: 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 5 - 2007 - Paper 1

Question 5

$Given-that-$y-=-3x^2-+-4-oot{x},-\,-x->-0$,-find--(a)-$-\frac{dy}{dx}-$-Edexcel-A-Level Maths Pure-Question 5-2007-Paper 1.png$

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 5 - 2007 - Paper 1

Step 1

96%

114 rated

Answer

To find ( \frac{dy}{dx} ), we need to differentiate the function:

\frac{dy}{dx} = \frac{d}{dx}(3x^2 + 4x^{\frac{1}{2}}) = 6x + 4 \cdot \frac{1}{2} x^{-\frac{1}{2}} = 6x + \frac{2}{\sqrt{x}}.

Thus, the derivative is:

\frac{dy}{dx} = 6x + \frac{2}{\sqrt{x}}.

Step 2

99%

104 rated

Answer

Now we differentiate ( \frac{dy}{dx} ) to find ( \frac{d^2y}{dx^2} ):

\frac{d^2y}{dx^2} = \frac{d}{dx} \left( 6x + \frac{2}{\sqrt{x}} \right) = 6 - \frac{2}{2} x^{-\frac{3}{2}} = 6 - x^{-\frac{3}{2}}.

So the second derivative is:

\frac{d^2y}{dx^2} = 6 - \frac{1}{x^{\frac{3}{2}}}.

Step 3

96%

101 rated

Answer

To integrate ( y ), we find:

\int y \, dx = \int (3x^2 + 4\sqrt{x}) \, dx.

Calculating this:

= \int 3x^2 \, dx + \int 4x^{\frac{1}{2}} \, dx = 3 \cdot \frac{x^3}{3} + 4 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + C = x^3 + \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.