How do you differentiate $F (t) = ln [\frac{{(2 t + 1)}^{3}}{{(3 t - 1)}^{4}}]$ $F(t)=ln[(2t+1)^3/(3t-1)^4]$ ?

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Answer 1 · 2016-08-08T14:33:33

$= - \frac{6 (t + 3)}{(2 t + 1) (3 t - 1)}$ $=- (6(t+ 3))/((2t+1)(3t-1))$

$F (t) = ln [\frac{{(2 t + 1)}^{3}}{{(3 t - 1)}^{4}}]$ $F(t)=ln[(2t+1)^3/(3t-1)^4]$

break it up to make it easier....
$= {ln (2 t + 1)}^{3} - {ln (3 t - 1)}^{4}$ $=ln(2t+1)^3 - ln (3t-1)^4$

$= 3 ln (2 t + 1) - 4 ln (3 t - 1)$ $=3ln(2t+1) - 4ln (3t-1)$

$\frac{d}{d t} (3 ln (2 t + 1) - 4 ln (3 t - 1))$ $d/dt (3ln(2t+1) - 4ln (3t-1))$

$= \frac{6}{2 t + 1} - \frac{12}{3 t - 1}$ $=6/(2t+1) - 12/(3t-1)$

$= \frac{18 t - 6 - 24 t - 12}{(2 t + 1) (3 t - 1)}$ $=(18t - 6 - 24t -12)/((2t+1)(3t-1))$

$= \frac{- 6 t - 18}{(2 t + 1) (3 t - 1)}$ $=(-6t - 18)/((2t+1)(3t-1))$

$= - \frac{6 (t + 3)}{(2 t + 1) (3 t - 1)}$ $=- (6(t+ 3))/((2t+1)(3t-1))$

How do you differentiate F(t)=ln[(2t+1)^3/(3t-1)^4]F(t)=ln[(2t+1)3(3t−1)4]F(t)=ln[(2t+1)^3/(3t-1)^4]?