C3 June 2012 Q7

Notes

Note that this is marked B1M1A1 on EPEN

(a)(i) M1 Attempts to differentiate ln(3x) to \(\dfrac{B}{x}\). Note that \(\dfrac{1}{3x}\) is fine.

M1 Attempts the product rule for \(x^{\frac{1}{2}}\ln(3x)\). If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms.
If the rule is not quoted nor implied from their stating of u, u’, v, v’ and their subsequent expression, only accept answers of the form
\(\ln(3x) \times Ax^{-\frac{1}{2}} + x^{\frac{1}{2}} \times \dfrac{B}{x}, \quad A, B \gt 0\)

A1 Any correct (un simplified) form of the answer. Remember to isw any incorrect further work
\(\dfrac{d}{dx}\left(x^{\frac{1}{2}}\ln(3x)\right) = \ln(3x) \times \dfrac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}} \times \dfrac{3}{3x} = \left(\dfrac{\ln(3x)}{2\sqrt{x}} + \dfrac{1}{\sqrt{x}}\right) = x^{-\frac{1}{2}}\left(\dfrac{1}{2}\ln 3x + 1\right)\)
Note that this part does not require the answer to be in its simplest form

(ii) M1 Applies the quotient rule, a version of which appears in the formula booklet. If the formula is quoted it must be correct. There must have been an attempt to differentiate both terms. If the formula is not quoted nor implied from their stating of u, u’, v, v’ and their subsequent expression, only accept answers of the form
\(\dfrac{(2x - 1)^5 \times \pm 10 - (1 - 10x) \times C(2x - 1)^4}{(2x - 1)^{10 \text{ or } 7 \text{ or } 25}}\)

A1 Any un simplified form of the answer. Eg \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(2x - 1)^5 \times -10 - (1 - 10x) \times 5(2x - 1)^4 \times 2}{((2x - 1)^5)^2}\)

A1 Cao. It must be simplified as required in the question \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{80x}{(2x - 1)^6}\)

(a)(ii) Alt using the product rule

Notes

M1 Knows that \(3\tan 2y\) differentiates to \(C\sec^2 2y\). The lhs can be ignored for this mark. If they write \(3\tan 2y\) as \(\dfrac{3\sin 2y}{\cos 2y}\) this mark is awarded for a correct attempt of the quotient rule.

A1 Writes down \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 6\sec^2 2y\) or implicitly to get \(1 = 6\sec^2 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Accept from the quotient rule \(\dfrac{6}{\cos^2 2y}\) or even \(\dfrac{\cos 2y \times 6\cos 2y - 3\sin 2y \times -2\sin 2y}{\cos^2 2y}\)

M1 An attempt to invert ‘their’ \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) to reach \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{f}(y)\), or changes the subject of their implicit differential to achieve a similar result \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{f}(y)\)

M1 Replaces an expression for \(\sec^2 2y\) in their \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) with \(x\) by attempting to use \(\sec^2 2y = 1 + \tan^2 2y\). Alternatively, replaces an expression for y in \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) with \(\dfrac{1}{2}\arctan\left(\dfrac{x}{3}\right)\)

A1 Any correct form of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of x. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{6\left(1 + \left(\frac{x}{3}\right)^2\right)}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{18 + 2x^2}\) or \(\dfrac{1}{6\sec^2\left(\arctan\left(\frac{x}{3}\right)\right)}\)

(b) Alternative using arctan. They must attempt to differentiate to score any marks. Technically this is M1A1M1A2