OCR MEI C3 2006 January — Question 1 4 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2006
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeShow that derivative equals expression
DifficultyModerate -0.8 This is a straightforward chain rule application with a simple power function. Students need to differentiate (1+6x)^(1/3) using the chain rule, then algebraically manipulate to show the result equals 2/y². The manipulation is routine: substitute y back in and simplify. This is easier than average as it's a single-technique question with minimal algebraic complexity, typical of an early C3 question designed to test basic chain rule competency.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
1 Given that \(y = ( 1 + 6 x ) ^ { \frac { 1 } { 3 } }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { y ^ { 2 } }\).
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(y = (1+6x)^{1/3} \Rightarrow \frac{dy}{dx} = \frac{1}{3}(1+6x)^{-2/3} \cdot 6\)M1 Chain rule
\(= \frac{1}{3}(1+6x)^{-2/3} \cdot 6\)B1 \(\frac{1}{3}(1+6x)^{-2/3}\) or \(\frac{1}{3}u^{-2/3}\)
\(= 2(1+6x)^{-2/3} = 2[(1+6x)^{1/3}]^{-2}\)A1 Any correct expression for the derivative
\(= \frac{2}{y^2}\) *E1 www
Or: \(y^3 = 1+6x \Rightarrow x = \frac{y^3-1}{6}\)M1 Finding \(x\) in terms of \(y\)
\(\frac{dx}{dy} = \frac{3y^2}{6} = \frac{y^2}{2}\)A1
\(\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{2}{y^2}\) *B1, E1 \(\frac{y^2}{2}\) o.e.
Or: \(y^3 = 1+6x \Rightarrow 3y^2\frac{dy}{dx} = 6\)M1 Together with attempt to differentiate implicitly
\(3y^2\frac{dy}{dx} = 6\)A1 \(3y^2\frac{dy}{dx} = 6\)
\(\frac{dy}{dx} = \frac{6}{3y^2} = \frac{2}{y^2}\) *A1, E1 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = (1+6x)^{1/3} \Rightarrow \frac{dy}{dx} = \frac{1}{3}(1+6x)^{-2/3} \cdot 6$ | M1 | Chain rule |
| $= \frac{1}{3}(1+6x)^{-2/3} \cdot 6$ | B1 | $\frac{1}{3}(1+6x)^{-2/3}$ or $\frac{1}{3}u^{-2/3}$ |
| $= 2(1+6x)^{-2/3} = 2[(1+6x)^{1/3}]^{-2}$ | A1 | Any correct expression for the derivative |
| $= \frac{2}{y^2}$ * | E1 | www |
| **Or:** $y^3 = 1+6x \Rightarrow x = \frac{y^3-1}{6}$ | M1 | Finding $x$ in terms of $y$ |
| $\frac{dx}{dy} = \frac{3y^2}{6} = \frac{y^2}{2}$ | A1 | |
| $\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{2}{y^2}$ * | B1, E1 | $\frac{y^2}{2}$ o.e. |
| **Or:** $y^3 = 1+6x \Rightarrow 3y^2\frac{dy}{dx} = 6$ | M1 | Together with attempt to differentiate implicitly |
| $3y^2\frac{dy}{dx} = 6$ | A1 | $3y^2\frac{dy}{dx} = 6$ |
| $\frac{dy}{dx} = \frac{6}{3y^2} = \frac{2}{y^2}$ * | A1, E1 | |

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1 Given that $y = ( 1 + 6 x ) ^ { \frac { 1 } { 3 } }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { y ^ { 2 } }$.

\hfill \mbox{\textit{OCR MEI C3 2006 Q1 [4]}}
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