Упр.28.28 ГДЗ Мордкович 10-11 класс (Алгебра)

а) y = (4х — 9)^7;
б) y = (x/3 + 2)^12;
в) у = (5х + 1)^9;
г) y = (x/4 — 3)^14.

1. $$y=(4x-9)^7$$
   $$y’ = 7(4x-9)^6 \cdot (4x-9)’ = 7(4x-9)^6 \cdot 4 = 28(4x-9)^6$$
2. $$y=\left(\frac{x}{3}+2\right)^{12}$$
   $$y’ = 12\left(\frac{x}{3}+2\right)^{11}\cdot \left(\frac{x}{3}+2\right)’ = 12\left(\frac{x}{3}+2\right)^{11}\cdot \frac13 = 4\left(\frac{x}{3}+2\right)^{11}$$
3. $$y=(5x+1)^9$$
   $$y’ = 9(5x+1)^8 \cdot (5x+1)’ = 9(5x+1)^8 \cdot 5 = 45(5x+1)^8$$
4. $$y=\left(\frac{x}{4}-3\right)^{14}$$
   $$y’ = 14\left(\frac{x}{4}-3\right)^{13}\cdot \left(\frac{x}{4}-3\right)’ = 14\left(\frac{x}{4}-3\right)^{13}\cdot \frac14 = \frac72\left(\frac{x}{4}-3\right)^{13}$$

Ответ

а) $$28(4x-9)^6$$
б) $$4\left(\frac{x}{3}+2\right)^{11}$$
в) $$45(5x+1)^8$$
г) $$\frac72\left(\frac{x}{4}-3\right)^{13}$$