The equation of a curve is $y=\frac{\ln 2x}{x}$ Find an expression for \(\frac{\mathrm{d}^2y}

Question

The equation of a curve is $y=\frac{\ln 2x}{x}$ Find an expression for $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ and show that $x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{x^2}=0$[5] Find the exact coordinates of the stationary point of the curve and use the second derivative test to determine if the stationary point is a maximum or minimum.[5]

Accepted Answer

(i) $\frac{d^2y}{dx^2}=\frac{2\ln(2x)-3}{x^3}$, and $x\frac{d^2y}{dx^2}+2\frac{dy}{dx}+\frac{1}{x^2}=0$ (ii) $\left(\frac{\mathrm{e}}{2},\frac{2}{\mathrm{e}}\right)$, maximum

2025 CHS PRELIMS P2 Q7

2025 CHS PRELIMS P2 Q7

Solution:

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