Question Number 27379 by abdo imad last updated on 05/Jan/18
$${let}\:{give}\:{f}\left({x}\right)=\:\:\int_{{x}} ^{\mathrm{2}{x}} \:\:\frac{{dt}}{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\:{calculate}\:{f}^{'} \left({x}\right). \\ $$
Commented by abdo imad last updated on 05/Jan/18
$${if}\:{f}\left({x}\right)=\:\int_{\alpha\left({x}\right)} ^{\beta\left({x}\right)\:\:} {u}\left({t}\right){dt}\:\:\:{with}\:\:\alpha\:{and}\:\beta\:{are}\:{are}\:{function}\:{of}\:{x} \\ $$$${f}^{'} \left({x}\right)=\:\beta^{'\left({x}\right)} {u}\left(\beta\left({x}\right)\right)−\alpha^{'} \left({x}\right)\:{u}\left(\alpha\left({x}\right)\right)\:{in}\:{the}\:{rxercise} \\ $$$${f}^{'} \left({x}\right)=\frac{\left(\mathrm{2}{x}\right)^{'} }{{ln}\left(\:\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)}\:−\:\frac{\left({x}\right)^{'} }{{ln}\left(\:\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$=\:\:\frac{\mathrm{2}}{{ln}\left(\:\mathrm{1}+\:\mathrm{4}{x}^{\mathrm{2}} \right)}\:−\frac{\mathrm{1}}{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\:. \\ $$
Answered by prakash jain last updated on 05/Jan/18
$${f}'\left({x}\right)=\frac{\mathrm{1}}{{ln}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)}\:\frac{{d}}{{dx}}\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\frac{{d}}{{dx}}{x}+\int_{{x}} ^{\:\mathrm{2}{x}} \mathrm{0}{dt} \\ $$$$=\frac{\mathrm{2}}{{ln}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)}−\frac{\mathrm{1}}{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$