Упр.27.15 ГДЗ Мордкович 10-11 класс (Алгебра)

Задача

выведите формулу дифференцирования функции:
а) у = х^2 + х;
б) у = 2х^2 — 3;
в) у = Зх — 2х^2;
г) у = х^4 + 4х — 5.

Подробный ответ

$$y=x^2+x$$
$$y(x+\Delta x)=(x+\Delta x)^2+(x+\Delta x)=x^2+2x\Delta x+(\Delta x)^2+x+\Delta x$$
$$\Delta y=y(x+\Delta x)-y(x)=2x\Delta x+(\Delta x)^2+\Delta x$$
$$\frac{\Delta y}{\Delta x}=2x+\Delta x+1$$
$$y'(x)=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0}(2x+\Delta x+1)=2x+1$$
$$y=2x^2-3$$
$$y(x+\Delta x)=2(x+\Delta x)^2-3=2x^2+4x\Delta x+2(\Delta x)^2-3$$
$$\Delta y=y(x+\Delta x)-y(x)=4x\Delta x+2(\Delta x)^2$$
$$\frac{\Delta y}{\Delta x}=4x+2\Delta x$$
$$y'(x)=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0}(4x+2\Delta x)=4x$$
$$y=3x-2x^2$$
$$y(x+\Delta x)=3(x+\Delta x)-2(x+\Delta x)^2=3x+3\Delta x-2x^2-4x\Delta x-2(\Delta x)^2$$
$$\Delta y=y(x+\Delta x)-y(x)=3\Delta x-4x\Delta x-2(\Delta x)^2$$
$$\frac{\Delta y}{\Delta x}=3-4x-2\Delta x$$
$$y'(x)=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0}(3-4x-2\Delta x)=3-4x$$
$$y=x^4+4x-5$$
$$y(x+\Delta x)=(x+\Delta x)^4+4(x+\Delta x)-5$$
$$y(x+\Delta x)=x^4+4x^3\Delta x+6x^2(\Delta x)^2+4x(\Delta x)^3+(\Delta x)^4+4x+4\Delta x-5$$
$$\Delta y=y(x+\Delta x)-y(x)=4x^3\Delta x+6x^2(\Delta x)^2+4x(\Delta x)^3+(\Delta x)^4+4\Delta x$$
$$\frac{\Delta y}{\Delta x}=4x^3+6x^2\Delta x+4x(\Delta x)^2+(\Delta x)^3+4$$
$$y'(x)=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0}\left(4x^3+6x^2\Delta x+4x(\Delta x)^2+(\Delta x)^3+4\right)=4x^3+4$$