OCR MEI C3 2011 January — Question 1 7 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2011
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeChain rule with single composition
DifficultyModerate -0.8 This is a straightforward application of the chain rule to differentiate a composite function. It requires rewriting the cube root as a power (1+x²)^(1/3), then applying the chain rule mechanically—a routine technique tested early in C3 with no problem-solving element.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

Given that \(y = \sqrt[3]{1 + x^2}\), find \(\frac{dy}{dx}\). [4]

AnswerMarks Guidance
\(y = \sqrt[3]{1 + x^2} = (1 + x^2)^{1/3}\)M1, M1, B1 Do not allow MR for square root their \(dy/dx \ne du/dx\) (available for wrong indices) no ft on \(\frac{1}{2}\) index
\(\frac{dy}{dx} = \frac{1}{3}(1 + x^2)^{-2/3} \cdot 2x\)
\(= \frac{2}{3}x(1 + x^2)^{-\frac{2}{3}}\)A1 [4] cao, mark final answer
$y = \sqrt[3]{1 + x^2} = (1 + x^2)^{1/3}$ | M1, M1, B1 | Do not allow MR for square root their $dy/dx \ne du/dx$ (available for wrong indices) no ft on $\frac{1}{2}$ index

$\frac{dy}{dx} = \frac{1}{3}(1 + x^2)^{-2/3} \cdot 2x$ | | 

$= \frac{2}{3}x(1 + x^2)^{-\frac{2}{3}}$ | A1 [4] | cao, mark final answer | oe e.g. $\frac{2x(1 + x^2)^{-\frac{2}{3}}}{3}$, $\frac{2x}{3\sqrt[3]{(1 + x^2)^2}}$, etc but must combine 2 with $1/3$

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Given that $y = \sqrt[3]{1 + x^2}$, find $\frac{dy}{dx}$.
[4]

\hfill \mbox{\textit{OCR MEI C3 2011 Q1 [7]}}