question\nfind the derivative of the function\n$y = 3x^{3}sqrt3{\frac{-2x - 4}{3x^{3}-6}}$ using logarithmic…

Answer:

$y'\left(x\right)=3x^{3}\sqrt[3]{\frac{-2x - 4}{3x^{3}-6}}\left(\frac{3}{x}-\frac{2}{3(-2x - 4)}-\frac{9x^{2}}{3(3x^{3}-6)}\right)$

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln\left(3x^{3}\right)+\frac{1}{3}\ln\left(\frac{-2x - 4}{3x^{3}-6}\right)$ $\ln y=\ln 3 + 3\ln x+\frac{1}{3}(\ln(-2x - 4)-\ln(3x^{3}-6))$

Step2: Differentiate both sides with respect to x

$\frac{y'}{y}=0+\frac{3}{x}+\frac{1}{3}\left(\frac{-2}{-2x - 4}-\frac{9x^{2}}{3x^{3}-6}\right)$

Step3: Solve for y'

$y' = y\left(\frac{3}{x}+\frac{1}{3}\left(\frac{-2}{-2x - 4}-\frac{9x^{2}}{3x^{3}-6}\right)\right)$ Substitute $y = 3x^{3}\sqrt[3]{\frac{-2x - 4}{3x^{3}-6}}$ into the above equation, we get $y'\left(x\right)=3x^{3}\sqrt[3]{\frac{-2x - 4}{3x^{3}-6}}\left(\frac{3}{x}-\frac{2}{3(-2x - 4)}-\frac{9x^{2}}{3(3x^{3}-6)}\right)$