Standard +0.3 This is a straightforward application of the quotient rule followed by solving dy/dx = 0 for a stationary point. The differentiation is routine (quotient rule with ln x), and finding the stationary point requires basic algebraic manipulation. Slightly above average difficulty only because it combines two standard techniques in sequence.
6 A curve has equation \(y = \frac { x } { 2 + 3 \ln x }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence find the exact coordinates of the stationary point of the curve.
When \(\frac{dy}{dx} = 0\): \(3\ln x - 1 = 0 \Rightarrow \ln x = 1/3 \Rightarrow x = e^{1/3}\)
Answer
Marks
Guidance
\(\Rightarrow y = \frac{e^{1/3}}{2 + 1} = \frac{1}{3}e^{1/3}\)
M1, B1, A1, M1, A1cao, M1, A1cao [7]
Quotient rule consistent with their derivatives or product rule + chain rule on \((2+3x)^{-1}\): \(\frac{d}{dx}(\ln x) = \frac{1}{x}\) soi; correct expression; their numerator \(= 0\) (or equivalent step from product rule formulation); M0 if denominator \(= 0\) is pursued; \(x = e^{1/3}\); substituting for their x (correctly); Must be exact: \(-0.46...\) is M1A0; www
$y = \frac{x}{2 + 3\ln x}$
$\Rightarrow \frac{dy}{dx} = \frac{(2 + 3\ln x) \cdot 1 - x \cdot \frac{3}{x}}{(2 + 3\ln x)^2} = \frac{2 + 3\ln x - 3}{(2 + 3\ln x)^2} = \frac{3\ln x - 1}{(2 + 3\ln x)^2}$
When $\frac{dy}{dx} = 0$: $3\ln x - 1 = 0 \Rightarrow \ln x = 1/3 \Rightarrow x = e^{1/3}$
$\Rightarrow y = \frac{e^{1/3}}{2 + 1} = \frac{1}{3}e^{1/3}$ | M1, B1, A1, M1, A1cao, M1, A1cao [7] | Quotient rule consistent with their derivatives or product rule + chain rule on $(2+3x)^{-1}$: $\frac{d}{dx}(\ln x) = \frac{1}{x}$ soi; correct expression; their numerator $= 0$ (or equivalent step from product rule formulation); M0 if denominator $= 0$ is pursued; $x = e^{1/3}$; substituting for their x (correctly); Must be exact: $-0.46...$ is M1A0; www
6 A curve has equation $y = \frac { x } { 2 + 3 \ln x }$. Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$. Hence find the exact coordinates of the stationary point of the curve.
\hfill \mbox{\textit{OCR MEI C3 2005 Q6 [7]}}