Ответ:

Применяем правила дифференцирования и таблицу производных .

[tex]\bf f(x)=\sqrt5\cdot x\ \ ,\ \ f'(x)=\sqrt5\\\\\\f(x)=\dfrac{3}{5x^{10}}\ \ ,\ \ f'(x)=\dfrac{0-3\cdot 5\cdot 10x^9}{25\, x^{20}}=-\dfrac{6}{x^{11}}\\\\\\f(x)=2x\sqrt[3]{\bf x}=3x^{^{\frac{4}{3}}}\ \ ,\ \ f'(x)=3\cdot \dfrac{4}{3}\cdot x^{^{\frac{1}{3}}}=4\sqrt[3]{x}\\\\\\f(x)=(\sqrt{x}+1)(3-2\sqrt{x})\ \ ,\ \ f'(x)=\dfrac{1}{2\sqrt{x}}\, (3-2\sqrt{x})+(\sqrt{x}+1)\cdot \dfrac{-2}{2\sqrt{x}}=\\\\=\dfrac{3}{2\sqrt{x}}-1-1-\dfrac{1}{\sqrt{x}}=\dfrac{1}{2\sqrt{x}}-2[/tex]

[tex]\bf f(x)=\dfrac{2x^2+3x}{x^2-4}\ \ ,\ \ f'(x)=\dfrac{(4x+3)(x^2-4)-(2x^2+3x)\cdot 2x}{(x^2-4)^2}=\\\\\\=\dfrac{4x^3-16x+3x^2-12-4x^3-6x^2}{(x^2-4)^2}=-\dfrac{3x^2+16x+12}{(x^2-4)^2}[/tex]