Aloha :)
zu 1)
$$\left(\ln\left(\frac{\overbrace{a+bx}^{=u}}{\underbrace{a-bx}_{=v}}\right)\right)'=\underbrace{\frac{1}{\frac{a+bx}{a-bx}}}_{\text{äußere}}\cdot\underbrace{\frac{\overbrace{b}^{=u'}\overbrace{(a-bx)}^{=v}-\overbrace{(a+bx)}^{=u}\overbrace{(-b)}^{=v'}}{\underbrace{(a-bx)^2}_{=v^2}}}_{\text{innere}}$$$$=\frac{a-bx}{a+bx}\cdot\frac{ab-b^2x+ab+b^2x}{(a-bx)^2}=\frac{2ab}{(a+bx)(a-bx)}=\frac{2ab}{a^2-b^2x^2}$$
zu 2)
$$\left(\frac{\overbrace{\ln x}^{=u}}{\underbrace{\sqrt{x^2-1}}_{=v}}\right)'=\frac{\overbrace{\frac{1}{x}}^{=u'}\overbrace{\sqrt{x^2-1}}^{=v}-\overbrace{\ln x}^{=u}\cdot\overbrace{\underbrace{\frac{1}{2\sqrt{x^2-1}}}_{\text{äußere}}\cdot\underbrace{2x}_{\text{innere}}}^{v'}}{\underbrace{x^2-1}_{=v^2}}=\frac{\frac{1}{x}\sqrt{x^2-1}-\frac{x\ln x}{\sqrt{x^2-1}}}{x^2-1}$$$$=\frac{\frac{1}{x}(x^2-1)-x\ln x}{(x^2-1)^{3/2}}=\frac{x^2-1-x^2\ln x}{x(x^2-1)^{3/2}}=\frac{x^2(1-\ln x)-1}{x(x^2-1)^{3/2}}$$
zu 3)
$$\left(\ln\sqrt{\frac{e^{2x}}{e^{2x}+1}}\right)'=\frac{1}{2}\left(\ln\left(\frac{e^{2x}}{e^{2x}+1}\right)\right)'=\frac{1}{2}\left(\ln\left(\frac{e^{2x}+1-1}{e^{2x}+1}\right)\right)'$$$$=\frac{1}{2}\left(\ln\left(1-\frac{1}{e^{2x}+1}\right)\right)'=\frac{1}{2}\underbrace{\frac{1}{1-\frac{1}{e^{2x}+1}}}_{=\text{äußere}}\cdot\underbrace{\left(-(e^{2x}+1)^{-1}\right)'}_{=\text{innere}}$$$$=\frac{1}{2}\frac{1}{\frac{e^{2x}}{e^{2x}+1}}\left(\underbrace{(e^{2x}+1)^{-2}}_{=\text{äußere}}\cdot \underbrace{e^{2x}}_{=\text{innere 1}}\cdot\underbrace{2}_{=\text{innere 2}}\right)=\frac{e^{2x}+1}{e^{2x}}\frac{1}{(e^{2x}+1)^2}e^{2x}$$$$=\frac{1}{e^{2x}+1}$$
