Функции сложные, поэтому немного теории перед нахождением их производных:
1. 
f (x) = 4x - \frac{2}{\sqrt{x}}\\ f' (x) = (4x - \frac{2}{\sqrt{x}})' = {4x}'-(\frac{2}{\sqrt{x}})'=4 - \frac{2' \cdot \sqrt{x}-2\cdot (\sqrt{x})'}{(\sqrt{x})^{2}}=\\ =4 - \frac{-2\cdot \frac{1}{2\sqrt{x}}}{(\sqrt{x})^{2}}=4 - (- \frac{\frac{1}{\sqrt{x}}}{(\sqrt{x})^{2}})=4 + \frac{1}{\sqrt{x}}:\frac{(\sqrt{x})^{2}}{1}=4+\frac{1}{\sqrt{x}}\cdot \frac{1}{x}=\\ =4+\frac{1}{\sqrt{x^{3}}}" alt="f (x) = 4x - \frac{2}{\sqrt{x}}\\ f' (x) = (4x - \frac{2}{\sqrt{x}})' = {4x}'-(\frac{2}{\sqrt{x}})'=4 - \frac{2' \cdot \sqrt{x}-2\cdot (\sqrt{x})'}{(\sqrt{x})^{2}}=\\ =4 - \frac{-2\cdot \frac{1}{2\sqrt{x}}}{(\sqrt{x})^{2}}=4 - (- \frac{\frac{1}{\sqrt{x}}}{(\sqrt{x})^{2}})=4 + \frac{1}{\sqrt{x}}:\frac{(\sqrt{x})^{2}}{1}=4+\frac{1}{\sqrt{x}}\cdot \frac{1}{x}=\\ =4+\frac{1}{\sqrt{x^{3}}}" align="absmiddle" class="latex-formula">
2. 