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f-x-3x-5-2x-1-f-x-




Question Number 198953 by ArifinTanjung last updated on 26/Oct/23
f(x)= ((3x−5)/(2x+1)) →f^′ (x)=....?
$$\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{3x}−\mathrm{5}}{\mathrm{2x}+\mathrm{1}}\:\rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=….? \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/23
f(x)= ((3x−5)/(2x+1)) →f^′ (x)=....?  ((u/v))^′ =((v u′−u v′)/v^2 )  u=3x−5⇒u′=3  v=2x+1⇒v′=2  f ′(x)=(((2x+1)(3)−(3x−5)(2))/((2x+1)^2 ))              =((6x+3−6x+10)/((2x+1)^2 ))              =((13)/((2x+1)^2 ))
$$\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{3x}−\mathrm{5}}{\mathrm{2x}+\mathrm{1}}\:\rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=….? \\ $$$$\left(\frac{{u}}{{v}}\right)^{'} =\frac{{v}\:{u}'−{u}\:{v}'}{{v}^{\mathrm{2}} } \\ $$$${u}=\mathrm{3}{x}−\mathrm{5}\Rightarrow{u}'=\mathrm{3} \\ $$$${v}=\mathrm{2}{x}+\mathrm{1}\Rightarrow{v}'=\mathrm{2} \\ $$$${f}\:'\left({x}\right)=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{3}\right)−\left(\mathrm{3}{x}−\mathrm{5}\right)\left(\mathrm{2}\right)}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{6}{x}+\mathrm{3}−\mathrm{6}{x}+\mathrm{10}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{13}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

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