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Question Number 127434 by MathSh last updated on 29/Dec/20

$$y=\frac{1}{4}ln\frac{1+x}{1-x}-\frac{1}{2}arctgx$$$$y^{\prime }=?$$

Commented by mohammad17 last updated on 29/Dec/20

$$y^{{}^{\prime }}=\frac{1}{4}\left(\frac{[(1-x)\left(1\right)-(1+x)(-1)](1+x)}{{(1-x)}^3}\right)-\frac{1}{2x^2+2}$$$$$$$$y^{{}^{\prime }}=\frac{2x+2}{{(1-x)}^3}-\frac{1}{2x^2+2}$$

Answered by Dwaipayan Shikari last updated on 29/Dec/20

$$y=\frac{1}{4}\left(log(1+x)\right)-\frac{1}{4}log(1-x)-\frac{1}{2}{tan}^{-1}x$$$$y^{\prime }=\frac{1}{4(1+x)}+\frac{1}{4(1-x)}-\frac{1}{2}.\frac{1}{1+x^2}=\frac{x^2}{1-x^4}$$

Answered by hknkrc46 last updated on 29/Dec/20

$$\bullet \frac{d}{dx}\ln \frac{a\left(x\right)}{b\left(x\right)}=\frac{a^{{}^{\prime }}\left(x\right)}{a\left(x\right)}-\frac{b^{{}^{\prime }}\left(x\right)}{b\left(x\right)}$$$$✓\frac{d}{dx}\ln \frac{1+x}{1-x}=\frac{{(1+x)}^{{}^{\prime }}}{1+x}-\frac{{(1-x)}^{\prime }}{1-x}$$$$=\frac{1}{1+x}+\frac{1}{1-x}$$$$\bullet \frac{d}{dx}\mathrm{arctg}x=\frac{1}{1+x^2}$$$$★y^{{}^{\prime }}=\frac{1}{4}(\frac{1}{1+x}+\frac{1}{1-x})-\frac{1}{2(1+x^2)}$$$$=\frac{1}{2}(\frac{1}{1-x^2}-\frac{1}{1+x^2})=\frac{x^2}{1-x^4}$$

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