Ответ:

Объяснение:

f(x) = 2x⁵ - x²/3 + 3x² - 4

[tex]f'(x) =5*2*x^4-2*x/3+2*3*x-0 =10*x^4+ 16x/3\\[/tex]

f(x) = (3x - 5)√x = 3x√x-5√x =

[tex]=3x^1^.^5 -5x^0^.^5 =3*1.5*x^0^.^5- 5*0.5*x^-^0^.^5=\\=4.5*\sqrt{x} -\frac{2.5}{\sqrt{x} }[/tex]

[tex]f(x)=\frac{x^2+9x}{x-4} = \frac{u}{v} u= x^2+9x ; v= x-4\\ f'(x) =\frac{u'*v-v'*u}{v^2} = \frac{(2x+9)*(x-4)-x^2-9x}{(x-4)^2} \\f'(x)=\frac{2x^2-8x+9x-36-x^2-9x}{(x-4)^2} =\frac{x^2-8x-36}{(x-4)^2}[/tex]