Jambmaths question:
$\frac{d}{dx}\left[ \log (4{{x}^{3}}-2x) \right]$ is equal to
Option A:
$\frac{12x-2}{4{{x}^{2}}}$
Option B:
$\frac{43{{x}^{2}}-2x}{7x}$
Option C:
$\frac{4{{x}^{2}}-2}{7x+6}$
Option D:
$\frac{12{{x}^{2}}-2}{4{{x}^{3}}-3x}$
Jamb Maths Solution:
$\begin{align} & \frac{d}{dx}\left[ \log (4{{x}^{3}}-2x) \right] \\ & \text{Note: }y=\log f(x),\text{ }\frac{dy}{dx}=\frac{f'(x)}{f(x)} \\ & \frac{d}{dx}\left[ \log (4{{x}^{3}}-2x) \right]=\frac{\tfrac{d}{dx}(4{{x}^{3}}-2x)}{4{{x}^{3}}-2x} \\ & \frac{dy}{dx}=\frac{12{{x}^{2}}-2}{4{{x}^{3}}-2x} \\\end{align}$
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