5 Given that \(y = ( 1 + 6 x ) ^ { \frac { 1 } { 3 } }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { y ^ { 2 } }\).
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Question 5:
Answer Marks
Guidance
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; \(\frac{1}{3}(1+6x)^{-2/3}\) or \(\frac{1}{3}u^{-2/3}\)
\(= 2(1+6x)^{-2/3}\) B1
Any correct expression for the derivative www
\(= 2[(1+6x)^{1/3}]^{-2}\) A1
\(= \frac{2}{y^2}\) E1
*Or*: \(y^3 = 1+6x \Rightarrow x = \frac{y^3-1}{6}\) M1
Finding \(x\) in terms of \(y\)
\(\Rightarrow \frac{dx}{dy} = \frac{3y^2}{6} = \frac{y^2}{2}\) A1 B1
\(y^2/2\) o.e.
\(\Rightarrow \frac{dy}{dx} = 1/(\frac{dx}{dy}) = \frac{2}{y^2}\) E1
*Or*: \(y^3 = 1+6x \Rightarrow 3y^2\frac{dy}{dx} = 6\) M1
Together with attempt to differentiate implicitly \(3y^2\frac{dy}{dx} = 6\)
\(\Rightarrow \frac{dy}{dx} = \frac{6}{3y^2} = \frac{2}{y^2}\) A1 A1 E1 [4]
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## Question 5:
| 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; $\frac{1}{3}(1+6x)^{-2/3}$ or $\frac{1}{3}u^{-2/3}$ |
| $= 2(1+6x)^{-2/3}$ | B1 | Any correct expression for the derivative www |
| $= 2[(1+6x)^{1/3}]^{-2}$ | A1 | |
| $= \frac{2}{y^2}$ | E1 | |
| *Or*: $y^3 = 1+6x \Rightarrow x = \frac{y^3-1}{6}$ | M1 | Finding $x$ in terms of $y$ |
| $\Rightarrow \frac{dx}{dy} = \frac{3y^2}{6} = \frac{y^2}{2}$ | A1 B1 | $y^2/2$ o.e. |
| $\Rightarrow \frac{dy}{dx} = 1/(\frac{dx}{dy}) = \frac{2}{y^2}$ | E1 | |
| *Or*: $y^3 = 1+6x \Rightarrow 3y^2\frac{dy}{dx} = 6$ | M1 | Together with attempt to differentiate implicitly $3y^2\frac{dy}{dx} = 6$ |
| $\Rightarrow \frac{dy}{dx} = \frac{6}{3y^2} = \frac{2}{y^2}$ | A1 A1 E1 [4] | |
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5 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 Q5 [4]}}