Edexcel C1 2010 January — Question 1 3 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2010
SessionJanuary
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind derivative of simple polynomial (integer powers)
DifficultyEasy -1.8 This is a straightforward application of the power rule for differentiation requiring only direct recall of a basic formula. It's a single-step C1 question with no problem-solving element, making it significantly easier than average A-level questions.
Spec1.07i Differentiate x^n: for rational n and sums

Given that \(y = x ^ { 4 } + x ^ { \frac { 1 } { 3 } } + 3\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^4 \to kx^3\) or \(x^{1/3} \to kx^{-2/3}\) or \(3 \to 0\) (\(k\) a non-zero constant)M1
\(\frac{dy}{dx} = 4x^3\ldots\) with '3' differentiated to zero (or 'vanishing')A1 1st A1 requires \(4x^3\), and 3 differentiated to zero; having '+C' loses the 1st A mark
\(\frac{dy}{dx} = \ldots + \frac{1}{3}x^{-2/3}\) or equivalent e.g. \(\frac{1}{3\sqrt[3]{x^2}}\) or \(\frac{1}{3(\sqrt[3]{x})^2}\)A1 Terms not added, but otherwise correct e.g. \(4x^3\), \(\frac{1}{3}x^{-2/3}\) loses the 2nd A mark
Total: [3]
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^4 \to kx^3$ or $x^{1/3} \to kx^{-2/3}$ or $3 \to 0$ ($k$ a non-zero constant) | M1 | |
| $\frac{dy}{dx} = 4x^3\ldots$ with '3' differentiated to zero (or 'vanishing') | A1 | 1st A1 requires $4x^3$, and 3 differentiated to zero; having '+C' loses the 1st A mark |
| $\frac{dy}{dx} = \ldots + \frac{1}{3}x^{-2/3}$ or equivalent e.g. $\frac{1}{3\sqrt[3]{x^2}}$ or $\frac{1}{3(\sqrt[3]{x})^2}$ | A1 | Terms not added, but otherwise correct e.g. $4x^3$, $\frac{1}{3}x^{-2/3}$ loses the 2nd A mark |

**Total: [3]**

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Given that $y = x ^ { 4 } + x ^ { \frac { 1 } { 3 } } + 3$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\

\hfill \mbox{\textit{Edexcel C1 2010 Q1 [3]}}