Question Number 197036 by SANOGO last updated on 06/Sep/23
$${calcule}\:{la}\:{derive}\:{de}: \\ $$$${g}\left({x}\right)=\:{arctan}\left(\frac{{x}−\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}\right) \\ $$
Answered by som(math1967) last updated on 06/Sep/23
$$\:{g}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{x}−\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}\right)^{\mathrm{2}} }×\frac{{d}}{{dx}}\left(\frac{{x}−\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}\right) \\ $$$$=\frac{\left(\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{2}} }{\left(\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{2}} +\left({x}−\mathrm{1}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{2}{x}−\mathrm{3}\right)−\mathrm{2}\left({x}−\mathrm{1}\right)}{\left(\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{10}} \\ $$