Question

Найдите производные:
a)y=x^4/4-4/x^4-8 √x
б)y=(x^2+1)cosx
в)y=x^2+3x/x-1

Answer 1

$a)y=\frac{x^4}4-\frac{4}{x^4}-8\sqrt{x}\\y'=x^3-4*(-4)*x^{-5}-8*\frac 12*x^{-\frac{1}2}=x^3+\frac{16}{x^5}-\frac{4}{\sqrt{x}}\\b)y=(x^2+1)cosx\\y'=2x*cosx-sinx(x^2+1)\\c)y=x^2+\frac{3x}{x-1}\\y'=2x+\frac{3(x-1)-3x}{(x-1)^2}=2x+\frac{3x-3-3x}{(x-1)^2}=2x-\frac{3}{(x-1)^2}$

Answer 2

$1)\\y=\frac{x^4}{4}-\frac{4}{x^4}-8\sqrt{x}\\\\y'=x^3+\frac{16}{x^5}-\frac{8}{2\sqrt{x}}\\\\\\2)\; y=(x^2+1)cosx\\\\y'=2xcosx-(x^2+1)sinx\\\\\\3)\; \; y=\frac{x^2+3x}{x-1}\\\\y'=\frac{(2x+3)(x-1)-(x^2+3x)\cdot 1}{(x-1)^2}=\frac{x^2-2x-3}{(x-1)^2}=\frac{(x+1)(x-3)}{(x-1)^2}$