Show that \(\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\ln 2x}{3x^2} \right) = \frac{1}{3x^3} -

Question

Show that $\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\ln 2x}{3x^2} \right) = \frac{1}{3x^3} - \frac{2\ln 2x}{3x^3}$.[3] Integrate $\frac{\ln 2x}{x^3}$ with respect to $x$.[3] Given that the curve $y = \mathrm{f}(x)$ passes through the point $ \left(\frac{1}{2}, 1 \right)$ and is such that $\mathrm{f}'(x) = \frac{\ln 2x}{x^3}$, find $\mathrm{f}(x)$.[3]

Accepted Answer

(a) $\frac{\mathrm d}{\mathrm dx}\left(\frac{\ln(2x)}{3x^2}\right)=\frac{1}{3x^3}-\frac{2\ln(2x)}{3x^3}$ (shown) (b) $-\frac{1}{4x^2}-\frac{\ln(2x)}{2x^2}+C$ (c) $f(x)=2-\frac{1}{4x^2}-\frac{\ln(2x)}{2x^2}$

2024 CGSS P1 Q12

2024 CGSS P1 Q12

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