Question Number 67960 by aseer imad last updated on 02/Sep/19
![(d/dx)[tan^(−1) ((4x)/( (√(1−4x^2 ))))] or (d/dx)tan^(−1) (2tanθ) [where 2x=sinθ ] which comes later if done considering 2x=sinθ please help](https://www.tinkutara.com/question/Q67960.png)
$$\frac{{d}}{{dx}}\left[{tan}^{−\mathrm{1}} \frac{\mathrm{4}{x}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}\right] \\ $$$${or} \\ $$$$\frac{{d}}{{dx}}{tan}^{−\mathrm{1}} \left(\mathrm{2}{tan}\theta\right)\:\:\:\:\:\:\:\left[{where}\:\mathrm{2}{x}={sin}\theta\:\right] \\ $$$$\:\:\:{which}\:{comes}\:{later}\:{if}\:{done}\:{considering} \\ $$$$\mathrm{2}{x}={sin}\theta \\ $$$${please}\:{help} \\ $$
Commented by Prithwish sen last updated on 02/Sep/19
$$\mathrm{Let}\:\mathrm{y}\:=\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4x}}{\:\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }} \\ $$$$\mathrm{Now}\:\mathrm{considering}\:\:\mathrm{2x}\:=\:\mathrm{sin}\theta \\ $$$$\mathrm{diff}.\:\mathrm{w}.\mathrm{r}.\mathrm{t}\:\theta\:\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{dx}}{\mathrm{d}\theta}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Cos}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }\: \\ $$$$\mathrm{Now}\:\mathrm{the}\:\mathrm{original}\:\mathrm{expression}\:\mathrm{turns}\:\mathrm{out} \\ $$$$\mathrm{y}=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2tan}\theta\right)\:\mathrm{again}\:\mathrm{diff}.\:\mathrm{w}.\mathrm{r}.\mathrm{t}\:\:\theta \\ $$$$\frac{\mathrm{dy}}{\mathrm{d}\theta}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4tan}^{\mathrm{2}} \theta}.\mathrm{2sec}^{\mathrm{2}} \theta\:=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{12x}^{\mathrm{2}} } \\ $$$$\therefore\frac{\boldsymbol{\mathrm{dy}}\:}{\boldsymbol{\mathrm{dx}}}\:=\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{d}\theta}}.\frac{\boldsymbol{\mathrm{d}\theta}}{\boldsymbol{\mathrm{dx}}}\:\:=\:\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{12}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }} \\ $$$$\mathrm{It}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{in}\:\mathrm{direct}\:\mathrm{method} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{4x}}{\:\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}\right)^{\mathrm{2}} }.\frac{\mathrm{4}\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }+\mathrm{4x}.\frac{\mathrm{8x}}{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}}{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} +\mathrm{16x}^{\mathrm{2}} }.\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{4x}^{\mathrm{2}} \right)+\mathrm{16x}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{4x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}\: \\ $$$$=\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{12}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
Commented by MJS last updated on 02/Sep/19
![(d/dx)[arctan u(x)]=((u′(x))/((u(x))^2 +1)) u(x)=((4x)/( (√(1−4x^2 )))) u′(x)=(4/( (√((1−4x^2 )^3 )))) (u(x))^2 +1=((12x^2 +1)/(1−4x^2 )) ((u′(x))/((u(x))^2 +1))=(4/((12x^2 +1)(√(1−4x^2 ))))](https://www.tinkutara.com/question/Q67968.png)
$$\frac{{d}}{{dx}}\left[\mathrm{arctan}\:{u}\left({x}\right)\right]=\frac{{u}'\left({x}\right)}{\left({u}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${u}\left({x}\right)=\frac{\mathrm{4}{x}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }} \\ $$$${u}'\left({x}\right)=\frac{\mathrm{4}}{\:\sqrt{\left(\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{3}} }} \\ $$$$\left({u}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{12}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$\frac{{u}'\left({x}\right)}{\left({u}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{4}}{\left(\mathrm{12}{x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }} \\ $$