Aloha :)
Deine erstes Frage hier? Na dann... Willkommen in der Mathelounge... \o/
$$f'_1(x)=\left(\frac{x^2}{\sqrt[3]{ax}}\right)'=\left(\frac{x^2}{\sqrt[3]{a}\cdot\sqrt[3]{x}}\right)'=\left(\frac{x^2}{\sqrt[3]{a}\cdot x^{\frac{1}{3}}}\right)'=\frac{1}{\sqrt[3]{a}}\left(x^{\frac{5}{3}}\right)'=\frac{1}{\sqrt[3]{a}}\cdot\frac{5}{3}\,x^{\frac{2}{3}}$$$$\phantom{f'_1(x)}=\frac{5}{3}\frac{1}{\sqrt[3]{a}}\frac{x}{x^{\frac{1}{3}}}=\frac{5}{3}\frac{1}{\sqrt[3]{a}}\frac{x}{\sqrt[3]x}=\frac{5x}{3\sqrt[3]{ax}}$$
$$f'_2(x)=\left(\sqrt{\sin(ax-c)}\right)'=\underbrace{\frac{1}{2\sqrt{\sin(ax-c)}}}_{=\text{äußere Abl.}}\cdot\underbrace{\overbrace{\cos(ax-c)}^{=\text{äußere}}\cdot \overbrace{a}^{=\text{innere}}}_{=\text{innere Abl.}}=\frac{a\cos(ax-c)}{2\sqrt{\sin(ax-c)}}$$
$$f'_3(x)=(\,\underbrace{e^{ax}}_{=u}\,\underbrace{(x^2+\tan x)}_{=v}\,)'=\underbrace{a\cdot e^{ax}}_{=u'}\,\underbrace{(x^2+\tan x)}_{=v}+\underbrace{e^{ax}}_{=u}\,\underbrace{(2x+(1+\tan^2x))}_{=v'}$$$$\phantom{f'_3(x)}=e^{ax}\left(a(x^2+\tan x)+2x+1+\tan^2x\right)=e^{ax}(ax^2+2x+1+\tan^2x+a\tan x)$$
$$f'_4(x)=\left((1+x^2)^x\right)'=\left(e^{x\ln(1+x^2)}\right)'=\underbrace{e^{x\ln(1+x^2)}}_{=\text{äußere}}\cdot\left(\underbrace{\overbrace{1}^{=u'}\cdot\overbrace{\ln(1+x^2)}^{=v}+\overbrace{x}^{=u}\cdot\overbrace{\frac{2x}{1+x^2}}^{=v'}}_{=\text{innere}}\right)$$$$\phantom{f'_4(x)}=(1+x^2)^x\cdot\left(\ln(1+x^2)+\frac{2x^2}{1+x^2}\right)$$
