Помогите решить 164 номер

Решите задачу:

$1)\quad f(x)=x\sqrt{x}=x\cdot x^{\frac{1}{2}} =x^{\frac{3}{2}}\\\\f'(x)=\frac{3}{2}\cdot x^{\frac{1}{2}}\\\\2)\quad f(x)=x^{\sqrt3}\\\\f'(x)=\sqrt3\cdot x^{\sqrt3-1}\\\\3)\quad f(x)= \frac{1}{x^{\frac{3}{4}}} =x^{-\frac{3}{4}}\\\\f'(x)=- \frac{3}{4} \cdot x^{-\frac{7}{4} }=-\frac{3}{4\cdot x^{\frac{7}{4}}}\\\\4)\quad f(x)= \frac{3}{\sqrt[3]{x}}=3\cdot x^{-\frac{1}{3}}\\\\f'(x)=-3\cdot \frac{1}{3}\cdot x^{-\frac{4}{3}}=-\frac{1}{x^{\frac{4}{3}}}\\\\5)\quad f(x)=x\sqrt[3]{x^2}=x\cdot x^{2/3}=x^{5/3}$

$f'(x)= \frac{5}{3}\cdot x^{\frac{2}{3}} \\\\6)\quad f(x)= \frac{x+5}{x^4} \\\\f'(x)= \frac{(x+5)'\cdot x^4-(x+5)\cdot (x^4)'}{(x^4)^2} =\frac{x^4-4x^3(x+5)}{x^8} = \frac{-3x-20}{x^5}$