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Question Number 67960 by aseer imad last updated on 02/Sep/19

$$\frac{d}{dx}\left[{tan}^{-1}\frac{4x}{\sqrt{1-4x^2}}\right]$$$$or$$$$\frac{d}{dx}{tan}^{-1}\left(2tan\theta \right)[where2x=sin\theta ]$$$$whichcomeslaterifdoneconsidering$$$$2x=sin\theta$$$$pleasehelp$$

Commented by Prithwish sen last updated on 02/Sep/19

$$\mathrm{Let}\mathrm{y}={\tan }^{-1}\frac{4\mathrm{x}}{\sqrt{1-{4\mathrm{x}}^2}}$$$$\mathrm{Now}\mathrm{considering}2\mathrm{x}=\sin \theta$$$$\mathrm{diff}.\mathrm{w}.\mathrm{r}.\mathrm{t}\theta \mathrm{we}\mathrm{get}\frac{\mathrm{dx}}{\mathrm{d}\theta }=\frac{1}{2}\mathrm{Cos}\theta =\frac{1}{2}\sqrt{1-{4\mathrm{x}}^2}$$$$\mathrm{Now}\mathrm{the}\mathrm{original}\mathrm{expression}\mathrm{turns}\mathrm{out}$$$$\mathrm{y}={\tan }^{-1}\left(2\tan \theta \right)\mathrm{again}\mathrm{diff}.\mathrm{w}.\mathrm{r}.\mathrm{t}\theta$$$$\frac{\mathrm{dy}}{\mathrm{d}\theta }=\frac{1}{1+{4\tan }^2\theta }.{2\sec }^2\theta =\frac{2}{1+{12\mathrm{x}}^2}$$$$\therefore \frac{\mathbf{dy}}{\mathbf{dx}}=\frac{\mathbf{dy}}{\mathbf{d}\mathit{\theta }}.\frac{\mathbf{d}\mathit{\theta }}{\mathbf{dx}}=\frac{4}{(1+12{\mathbf{x}}^2)\sqrt{1-4{\mathbf{x}}^2}}$$$$\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{done}\mathrm{in}\mathrm{direct}\mathrm{method}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+{\left(\frac{4\mathrm{x}}{\sqrt{1-{4\mathrm{x}}^2}}\right)}^2}.\frac{4\sqrt{1-{4\mathrm{x}}^2}+4\mathrm{x}.\frac{8\mathrm{x}}{2\sqrt{1-{4\mathrm{x}}^2}}}{1-{4\mathrm{x}}^2}$$$$=\frac{1-{4\mathrm{x}}^2}{1-{4\mathrm{x}}^2+{16\mathrm{x}}^2}.\frac{4(1-{4\mathrm{x}}^2)+{16\mathrm{x}}^2}{(1-{4\mathrm{x}}^2)\sqrt{1-{4\mathrm{x}}^2}}$$$$=\frac{4}{(1+12{\mathbf{x}}^2)\sqrt{1-4{\mathbf{x}}^2}}\mathbf{please}\mathbf{check}.$$

Commented by MJS last updated on 02/Sep/19

$$\frac{d}{dx}\left[\arctan u\left(x\right)\right]=\frac{u^{\prime }\left(x\right)}{{\left(u\left(x\right)\right)}^2+1}$$$$u\left(x\right)=\frac{4x}{\sqrt{1-4x^2}}$$$$u^{\prime }\left(x\right)=\frac{4}{\sqrt{{(1-4x^2)}^3}}$$$${\left(u\left(x\right)\right)}^2+1=\frac{12x^2+1}{1-4x^2}$$$$\frac{u^{\prime }\left(x\right)}{{\left(u\left(x\right)\right)}^2+1}=\frac{4}{(12x^2+1)\sqrt{1-4x^2}}$$

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