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$\displaystyle \left[\ln\left(\dfrac{1+x}{1-x}\right)\right]'$

$\displaystyle \left[\ln\left(\dfrac{1+x}{1-x}\right)\right]'= \dfrac{1-x}{1+x}\left[\dfrac{1+x}{1-x}\right]'= \dfrac{1-x}{1+x}\dfrac{[1+x]'(1-x)-(1+x)[1-x]'}{(1-x)^2}= \dfrac{1-x}{1+x}\dfrac{(1-x)+(1+x)}{(1-x)^2}= \bbox[#FFECB3,5px]{\dfrac{2}{(1+x)(1-x)}}$

Alternatively,

$\displaystyle \left[\ln\left(\dfrac{1+x}{1-x}\right)\right]'= [\ln(1+x)-\ln(1+x)]'= \dfrac{1}{1+x}[1+x]'-\dfrac{1}{1-x}[1-x]'= \dfrac{1}{1+x}+\dfrac{1}{1-x}= \dfrac{(1-x)+(1+x)}{(1+x)(1-x)}= \bbox[#FFECB3,5px]{\dfrac{2}{(1+x)(1-x)}}$

$\displaystyle \left[\dfrac{x\sin(2x)}{\tan(3x)}\right]'$

$\displaystyle \left[\dfrac{x\sin(2x)}{\tan(3x)}\right]'= \dfrac{[x\sin(2x)]'(\tan(3x))-(x\sin(2x))[\tan(3x)]'}{(\tan(3x))^2}= \dfrac{(\sin(2x)+2x\cos(2x))(\tan(3x))-(x\sin(2x))\left(\frac{3}{\cos^2(3x)}\right)}{\tan^2(3x)}=\\ \qquad\displaystyle \bbox[#FFECB3,5px]{\dfrac{\sin(2x)+2x\cos(2x)}{\tan(3x)}-\dfrac{3x\sin(2x)}{\sin^2(3x)}}$

$\displaystyle \left[\sin\left(\dfrac{x}{x+\sin x}\right)\right]'$

$\displaystyle \left[\sin\left(\dfrac{x}{x+\sin x}\right)\right]'= \cos\left(\dfrac{x}{x+\sin x}\right)\left[\dfrac{x}{x+\sin x}\right]'= \cos\left(\dfrac{x}{x+\sin x}\right)\dfrac{[x]'(x+\sin x)-x[x+\sin x]'}{(x+\sin x)^2}=\\ \qquad\displaystyle \cos\left(\dfrac{x}{x+\sin x}\right)\dfrac{(x+\sin x)-x(1+\cos x)}{(x+\sin x)^2}= \bbox[#FFECB3,5px]{\cos\left(\dfrac{x}{x+\sin x}\right)\dfrac{\sin x-x\cos x}{(x+\sin x)^2}}$

$\displaystyle \left[\cos^3(1-x^2)\right]'$

$\displaystyle \left[\cos^3(1-x^2)\right]'= 3\cos^2(1-x^2)[\cos(1-x^2)]'= -3\cos^2(1-x^2)\sin(1-x^2)[1-x^2]= \bbox[#FFECB3,5px]{6x\cos^2(1-x^2)\sin(1-x^2)}$

$\displaystyle \left[\sqrt[3]{x+\sqrt{3x}}\right]'$

$\displaystyle \left[\sqrt[3]{x+\sqrt{3x}}\right]'= (x+\sqrt{3x})^{\frac{1}{3}}= \dfrac{1}{3}(x+\sqrt{3x})^{-\frac{2}{3}}[x+\sqrt{3x}]'= \bbox[#FFECB3,5px]{\dfrac{1}{3}(x+\sqrt{3x})^{-\frac{2}{3}}\left(1+\dfrac{\sqrt{3}}{2\sqrt{x}}\right)}$