Question Number 186726 by Rupesh123 last updated on 09/Feb/23
Answered by cortano12 last updated on 09/Feb/23
$${f}\:'\left({x}\right)=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{ln}\:{x}^{\mathrm{3}} }=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{ln}\:{ex}^{\mathrm{3}} } \\ $$$${f}\:''\left({x}\right)=\:\frac{\mathrm{6}{x}.\mathrm{ln}\:{ex}^{\mathrm{3}} −\frac{\mathrm{3}{ex}^{\mathrm{2}} }{{ex}^{\mathrm{3}} }\:.\mathrm{3}{x}^{\mathrm{2}} }{\left(\mathrm{ln}\:{ex}^{\mathrm{3}} \right)^{\mathrm{2}} } \\ $$
Answered by Ar Brandon last updated on 09/Feb/23
![(∂/∂t)∫_(u(t)) ^(v(t)) f(x, t)dx=∫_(u(t)) ^(v(t)) (∂/∂t)f(x, t)dx+[f(v(t), t)]v′(t)−[f(u(t), t)]u′(t)](https://www.tinkutara.com/question/Q186731.png)
$$\frac{\partial}{\partial{t}}\int_{{u}\left({t}\right)} ^{\mathrm{v}\left({t}\right)} {f}\left({x},\:{t}\right){dx}=\int_{{u}\left({t}\right)} ^{\mathrm{v}\left({t}\right)} \frac{\partial}{\partial{t}}{f}\left({x},\:{t}\right){dx}+\left[{f}\left(\mathrm{v}\left({t}\right),\:{t}\right)\right]\mathrm{v}'\left({t}\right)−\left[{f}\left({u}\left({t}\right),\:{t}\right)\right]{u}'\left({t}\right) \\ $$