Упр.7.3 ГДЗ Мордкович Семенов 11 класс (Алгебра)

а) y=2x^2; в) y=-(2/3)x^2;
б) y=x^2-x; г) y=x^2+3x.

1. $$y=2x^2$$

   $$y(x+\Delta x)=2(x+\Delta x)^2=2x^2+4x\Delta x+2(\Delta x)^2$$
   $$\Delta y=y(x+\Delta x)-y(x)=4x\Delta x+2(\Delta x)^2$$
   $$y'(x)=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} =\lim_{\Delta x\to 0}(4x+2\Delta x)=4x$$

2. $$y=x^2-x$$

   $$y(x+\Delta x)=(x+\Delta x)^2-(x+\Delta x)=x^2+2x\Delta x+(\Delta x)^2-x-\Delta x$$
   $$\Delta y=2x\Delta x+(\Delta x)^2-\Delta x$$
   $$y'(x)=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} =\lim_{\Delta x\to 0}(2x+\Delta x-1)=2x-1$$

3. $$y=-\frac{2}{3}x^2$$

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

4. $$y=x^2+3x$$

   $$y(x+\Delta x)=(x+\Delta x)^2+3(x+\Delta x)=x^2+2x\Delta x+(\Delta x)^2+3x+3\Delta x$$
   $$\Delta y=2x\Delta x+(\Delta x)^2+3\Delta x$$
   $$y'(x)=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} =\lim_{\Delta x\to 0}(2x+\Delta x+3)=2x+3$$

Ответ

$$a)\ y’=4x;\quad б)\ y’=2x-1;\quad в)\ y’=-\frac{4}{3}x;\quad г)\ y’=2x+3.$$