A-Level Edexcel Maths Pure Modelling with Functions: Differentiate with respect to $x

Differentiate with respect to $x$, giving each answer in its simplest form - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1

Question 9

Differentiate with respect to $x$, giving each answer in its simplest form. (a) $(1 - 2x)^2$ (b) $\frac{x^3 + 6\sqrt{x}}{2x^2}$

Worked Solution & Example Answer:Differentiate with respect to $x$, giving each answer in its simplest form - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1

Step 1

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Answer

To differentiate $(1 - 2x)^2$ , we will use the chain rule.

Let $u = 1 - 2x$ , then $(1 - 2x)^2 = u^2$ .
Differentiate using the chain rule:

$\frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx}$

where $\frac{du}{dx} = -2$ .
Therefore,

$\frac{d}{dx}((1 - 2x)^2) = 2(1 - 2x)(-2) = -4(1 - 2x).$
Expanding gives:

$-4(1 - 2x) = -4 + 8x.$
The final answer is:

$-4 + 8x$

Step 2

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Answer

For this differentiation, we will use the quotient rule:

If $y = \frac{f(x)}{g(x)}$ , then $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ where:

Calculate $f'(x)$ :
- $f'(x) = 3x^2 + 6 \cdot \frac{1}{2} x^{-1/2} = 3x^2 + \frac{3}{\sqrt{x}}.$
Calculate $g'(x)$ :
- $g'(x) = 4x.$
Substitute into the quotient rule:

$\frac{dy}{dx} = \frac{(3x^2 + \frac{3}{\sqrt{x}})(2x^2) - (x^3 + 6\sqrt{x})(4x)}{(2x^2)^2}$
Expanding and simplifying the numerator:

$= \frac{6x^4 + 6x - 4x^4 - 24x^{3/2}}{4x^4}$

$= \frac{2x^4 + 6x - 24x^{3/2}}{4x^4}$

$= \frac{1}{2} + \frac{3}{2x^3} - \frac{6}{x^{5/2}}$ .
The final simplified answer is:

$\frac{1}{2} + \frac{3}{2x^3} - \frac{6}{x^{5/2}}$