Additional Math – Differentiate – Natural Log (ln) and Cube Root

$latex \displaystyle \text{y = ln}\sqrt[3]{{\frac{{3x-1}}{{x+5}}}}$

Before you Differentiate, simplify the expression by changing the cube root to index form followed by using the Logarithm Law  $latex \displaystyle {{\log }_{a}}{{x}^{n}}\ =n\times {{\log }_{a}}x$

$latex \displaystyle y=\ln {{(\frac{{3x-1}}{{x+5}})}^{{\frac{1}{3}}}}$

$latex \displaystyle y=\frac{1}{3}\ln (\frac{{3x-1}}{{x+5}})$

Next use the Logarithm Law  $latex \displaystyle {{\log }_{a}}(\frac{x}{y})\ =\log {}_{a}x-{{\log }_{a}}y$ to continue to simplify the expression before differentiating.

$latex \displaystyle y=\frac{1}{3}\left[ {\ln (3x-1)-\ln x+5)} \right]$

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{1}{3}\left[ {\frac{1}{{3x-1}}\times 3-\frac{1}{{x+5}}\times 1} \right]$

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{1}{3}\left[ {\frac{3}{{3x-1}}-\frac{1}{{x+5}}} \right]$

$latex \displaystyle \frac{{dy}}{{dx}}=\frac{1}{{3x-1}}-\frac{1}{{3\left( {x+5} \right)}}$


As I said before in the previous post regarding the differentiation of log, It is easier to simplify the expression using law of logarithm before differentiation as you need fewer steps and you will less likely to make mistakes.


New Elementary Math (E-Math) and Additional Math Group  Tuition Class near Admiralty MRT station.

A-Math & E-Math Tuition Class at Woodlands and Johor Bahru.


More Posts