Answer
$g'(t)=\frac{45(t-2)^8}{(2t+1)^{10}}$
Work Step by Step
Given:
$g(t)=(\frac{t-2}{2t+1})^9$
Explanation:
Differentiate w.r.t "t".
$g'(t)=\frac{d}{dt}[(\frac{t-2}{2t+1})^9]$
Using power rule.
$=9(\frac{t-2}{2t+1})^{9-1}\frac{d}{dt}(\frac{t-2}{2t+1})$
$=9(\frac{t-2}{2t+1})^{8}\frac{d}{dt}(\frac{t-2}{2t+1})$
Applying quotient rule.
$=9(\frac{t-2}{2t+1})^8[{\frac{(2t+1)\frac{d}{dt}(t-2)-\frac{d}{dt}(2t+1)(t-2)}{(2t+1)^2}}]$
$=9(\frac{t-2}{2t+1})^8[{\frac{(2t+1)(\frac{d}{dt}(t)-\frac{d}{dt}(2))-\frac{d}{dt}(t)+\frac{d}{dt}(1))(t-2)}{(2t+1)^2}}]$
$=9(\frac{t-2}{2t+1})^8[{\frac{(2t+1)(1)-(0)-2((1)+(0))(t-2)}{(2t+1)^2}}]$
$=9(\frac{t-2}{2t+1})^8[{\frac{(2t+1)-2(t-2)}{(2t+1)^2}}]$
$=9(\frac{t-2}{2t+1})^8[{\frac{2t+1-2t+4}{(2t+1)^2}}]$
$=9(\frac{t-2}{2t+1})^8({\frac{5}{(2t+1)^2}})$
$=45({\frac{t-2}{(2t+1)^2}})^8$
$=45({\frac{(t-2)^8}{(2t+1)^{10}}}$
so,
$g'(t)={\frac{45(t-2)^8}{(2t+1)^{10}}}$
