Moderate -0.5 This is a straightforward application of the first principles definition with a simple power term. While it requires algebraic manipulation of the difference quotient and factoring a cubic expression, the steps are mechanical and follow a standard template taught explicitly for polynomial differentiation from first principles. It's slightly easier than average because it's a single-term polynomial with a clean result, though not trivial since it requires careful algebra.
3. Given that
$$y = \frac { 1 } { 3 } x ^ { 3 }$$
use differentiation from first principle to show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 }$$
Begins the process by writing down the gradient of the chord and attempts to expand the correct bracket. Example: \(\frac{\frac{1}{3}(x+h)^3 - \frac{1}{3}x^3}{h} = \frac{\frac{1}{3}(x^3 + 3x^2h + 3xh^2 + h^3) - \frac{1}{3}x^3}{h} =\)
A1
Reaches a correct fraction or equivalent with the \(x^3\) terms cancelled out. Example: \(\frac{\frac{1}{3}(x^3 + 3x^2h + 3xh^2 + h^3) - \frac{1}{3}x^3}{h} = \frac{\frac{1}{3}(3x^2h + 3xh^2 + \frac{1}{3}h^3)}{h} = x^2 + xh + \frac{1}{3}h^2\)
A1
Completes the process by applying a limiting argument and deduces that \(\frac{dy}{dx} = x^2\) with no errors seen.
The " \(\frac{dy}{dx}\) " doesn't have to appear but there must be something equivalent e.g. "f'(x) =" or "Gradient =" which can appear anywhere in their working. If f'(x) is used then there is no requirement to see f(x) defined first. Condone e.g. \(\frac{dy}{dx} \to x^2\) or \(f'(x) \to x^2\)
but e.g. \(\frac{dy}{dx} = \frac{x^2h + xh^2 + \frac{1}{3}h^3}{h} = x^2 + x(0) + \frac{1}{3}(0)^2 = x^2\) is not if there is no \(h \to 0\) seen. The \(h \to 0\) does not need to be present throughout the proof. e.g. on every line.
They must reach \(x^2 + xh + \frac{1}{3}h^2\) at the end and not \(\frac{x^2h + xh^2 + \frac{1}{3}h^3}{h}\) to complete the limiting argument.
(3 marks)
**M1** | Begins the process by writing down the gradient of the chord and attempts to expand the correct bracket. Example: $\frac{\frac{1}{3}(x+h)^3 - \frac{1}{3}x^3}{h} = \frac{\frac{1}{3}(x^3 + 3x^2h + 3xh^2 + h^3) - \frac{1}{3}x^3}{h} =$
**A1** | Reaches a correct fraction or equivalent with the $x^3$ terms cancelled out. Example: $\frac{\frac{1}{3}(x^3 + 3x^2h + 3xh^2 + h^3) - \frac{1}{3}x^3}{h} = \frac{\frac{1}{3}(3x^2h + 3xh^2 + \frac{1}{3}h^3)}{h} = x^2 + xh + \frac{1}{3}h^2$
**A1** | Completes the process by applying a limiting argument and deduces that $\frac{dy}{dx} = x^2$ with no errors seen.
The " $\frac{dy}{dx}$ " doesn't have to appear but there must be something equivalent e.g. "f'(x) =" or "Gradient =" which can appear anywhere in their working. If f'(x) is used then there is no requirement to see f(x) defined first. Condone e.g. $\frac{dy}{dx} \to x^2$ or $f'(x) \to x^2$
Condone missing brackets so allow e.g. $\frac{dy}{dx} = \lim_{h \to 0} \frac{x^2h + xh^2 + \frac{1}{3}h^3}{h} = \lim_{h \to 0} x^2 + xh + \frac{1}{3}h^2 = x^2$
Do not allow $h = 0$ if there is never a reference to $h \to 0$
Example: $\frac{dy}{dx} = \lim_{h \to 0} \frac{x^2h + xh^2 + \frac{1}{3}h^3}{h} = \lim_{h \to 0} x(0) + \frac{1}{3}(0)^2 = x^2$ is acceptable
but e.g. $\frac{dy}{dx} = \frac{x^2h + xh^2 + \frac{1}{3}h^3}{h} = x^2 + x(0) + \frac{1}{3}(0)^2 = x^2$ is not if there is no $h \to 0$ seen. The $h \to 0$ does not need to be present throughout the proof. e.g. on every line.
They must reach $x^2 + xh + \frac{1}{3}h^2$ at the end and not $\frac{x^2h + xh^2 + \frac{1}{3}h^3}{h}$ to complete the limiting argument.
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3. Given that
$$y = \frac { 1 } { 3 } x ^ { 3 }$$
use differentiation from first principle to show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 }$$
\hfill \mbox{\textit{Edexcel PMT Mocks Q3 [3]}}