A-Level Edexcel Maths Pure Applications of Differentiation: 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 can apply the chain rule.

Let $u = 1 - 2x$ , then $y = u^2$ .
Differentiate $y$ with respect to $u$ : $\frac{dy}{du} = 2u$
Differentiate $u$ with respect to $x$ : $\frac{du}{dx} = -2$
Now apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2(1 - 2x)(-2)$
Simplifying gives: $\frac{dy}{dx} = -4(1 - 2x)$

Step 2

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Answer

For this part, we will use the quotient rule to differentiate.

Let $f(x) = x^3 + 6\sqrt{x}$ and $g(x) = 2x^2$ .
The quotient rule states: $\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f' g - f g'}{g^2}$
First, find $f'$ $f^{'}$ and $g'$ $g^{'}$ :
- $f'(x) = 3x^2 + \frac{6}{2\sqrt{x}} = 3x^2 + \frac{3}{\sqrt{x}}$
- $g'(x) = 4x$
Now apply the quotient rule: $\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{(3x^2 + \frac{3}{\sqrt{x}})(2x^2) - (x^3 + 6\sqrt{x})(4x)}{(2x^2)^2}$
Simplifying the numerator and then dividing by $4x^4$ will give the final derivative in its simplest form.