If \( y=x^3\log(3x^2+6) \), then \( \displaystyle\frac{dy}{dx}= \)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{x^2}{x^2+2}[(3x^2+6)\log(3x^2+6)+2x^2] \)
By the product rule, with the chain rule for the logarithm,
\( \displaystyle \frac{dy}{dx}=3x^2\log(3x^2+6)+x^3\cdot\frac{6x}{3x^2+6} \)
\( \displaystyle =3x^2\log(3x^2+6)+\frac{6x^4}{3(x^2+2)} \)
\( \displaystyle =3x^2\log(3x^2+6)+\frac{2x^4}{x^2+2} \)
To match the form of the options, \( \frac{x^2}{x^2+2} \) is taken out as a common factor, using
\( \displaystyle 3x^2=\frac{x^2}{x^2+2}\cdot 3(x^2+2)=\frac{x^2}{x^2+2}(3x^2+6) \)
\( \displaystyle \frac{dy}{dx}=\frac{x^2}{x^2+2}\left[(3x^2+6)\log(3x^2+6)+2x^2\right] \)
Why the other options are incorrect
Option (1) \( \left(\displaystyle\frac{x^2+2}{x^2}\right)[(3x^2+6)\log(3x^2+6)+2x^2] \) — Incorrect. The factor is turned upside down; \( \frac{x^2+2}{x^2} \) is the reciprocal of the correct factor \( \frac{x^2}{x^2+2} \).
Option (2) \( \left(1+\displaystyle\frac{2}{x^2}\right)[(3x^2+6)\log(3x^2+6)+2x^2] \) — Incorrect. \( 1+\frac{2}{x^2}=\frac{x^2+2}{x^2} \), so this is the same expression as Option (1), with the factor upside down.
Option (4) \( \left(x^2+\displaystyle\frac{1}{2}\right)[(x^2+2)\log(3x^2+6)+2x^2] \) — Incorrect. At x = 1 this gives \( \frac{3}{2}(3\log 9+2) \), while the correct value is \( \frac{1}{3}(9\log 9+2)=3\log 9+\frac{2}{3} \).