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

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