Logarithmic differentiation of $y=(x^2(7x-14)^{1/3})/(1+x^2)^4$ Honestly I have no clue how to rewrite then start it. I know you have to Ln both sides but how would you Ln the right side?
Remember that $\ln$ turns products into sums, powers into products and divisions into subtractions: $$\begin{align}\ln y &amp;= \ln \left(\frac{x^2(7x-14)^{1/3}}{(1+x^2)^4}\right) \\ &amp;= \ln(x^{\color{red}{2}}) + \ln((7x-14)^{\color{blue}{1/3}}) - \ln((1+x^2)^{\color{green}{4}}) \\ &amp;= \color{red}{2}\ln x + \color{blue}{\frac{1}{3}}\ln(7x-14) - \color{green}{4}\ln(1+x^2). \end{align}$$
We know that for real numbers $a,b,c,d&gt;0$ : $$
\ln(a/b)=\ln(a)-\ln(b), \ln(cd)=\ln(c)+\ln(d)
$$
Let $a=x^2(7x-14)^{1/3},b=(1+x^2)^4$ and applying the two formulas.
Then differentiate. For the future, take care at your title question and consistence of it...$-1$.