\(\frac{d}{dx} [\log (4x^3 – 2x)]\) is equal to
- \(\frac{12x - 2}{4x^2}\)
- \(\frac{43x^2 - 2x}{7x}\)
- \(\frac{4x^2 - 2}{7x + 6}\)
-
\(\frac{12x^2 - 2}{4x^3 - 2x}\)
The correct answer is: D
Explanation
\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1)
Let u = 4x\(^3\) - 2x.
\(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
\(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\)
\(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\)
\(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\)
= \(\frac{12x^2 - 2}{4x^3 - 2x}\)
There is an explanation video available .