Упр.39.12 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

1) y=(2x+3)^5; 5) y=3ctg(x/5); 9) y=1/(4x+5);
2) y=(1/3 x-6)^18; 6) y=v(2x+1); 10) y=(x^2/2+4x-1)^(-6);
3) y=cos(2x); 7) y=(1-x)^(1/3); 11) y=vsin(x);
4) y=sin^2(x); 8) y=v(x^2+1); 12) y=sin(vx).

1. $$y=(2x+3)^5$$
   $$y’ = 5(2x+3)^4 \cdot 2 = 10(2x+3)^4.$$
2. $$y=\left(\frac13x-6\right)^{18}$$
   $$y’ = 18\left(\frac13x-6\right)^{17}\cdot \frac13 = 6\left(\frac13x-6\right)^{17}.$$
3. $$y=\cos 2x$$
   $$y’=-\sin 2x \cdot 2=-2\sin 2x.$$
4. $$y=\sin^2 x$$
   $$y’ = 2\sin x \cdot \cos x = \sin 2x.$$
5. $$y=3\ctg\frac{x}{5}$$
   $$y’ = 3\left(-\frac{1}{\sin^2\frac{x}{5}}\right)\cdot \frac15 = -\frac{3}{5\sin^2\frac{x}{5}}.$$
6. $$y=\sqrt{2x+1}$$
   $$y’=\frac{1}{2\sqrt{2x+1}}\cdot 2=\frac{1}{\sqrt{2x+1}}.$$
7. $$y=\sqrt[3]{1-x}=(1-x)^{\frac13}$$
   $$y’=\frac13(1-x)^{-\frac23}\cdot(-1)=-\frac{1}{3\sqrt[3]{(1-x)^2}}.$$
8. $$y=\sqrt{x^2+1}$$
   $$y’=\frac{1}{2\sqrt{x^2+1}}\cdot 2x=\frac{x}{\sqrt{x^2+1}}.$$
9. $$y=\frac{1}{4x+5}=(4x+5)^{-1}$$
   $$y’=-1\cdot(4x+5)^{-2}\cdot 4=-\frac{4}{(4x+5)^2}.$$
10. $$y=\left(\frac{x^2}{2}+4x-1\right)^{-6}$$
    $$y’=-6\left(\frac{x^2}{2}+4x-1\right)^{-7}\cdot\left(x+4\right)$$
    $$y’=-\frac{6x+24}{\left(\frac{x^2}{2}+4x-1\right)^7}.$$
11. $$y=\sqrt{\sin x}$$
    $$y’=\frac{1}{2\sqrt{\sin x}}\cdot \cos x=\frac{\cos x}{2\sqrt{\sin x}}.$$
12. $$y=\sin\sqrt{x}$$
    $$y’=\cos\sqrt{x}\cdot \frac{1}{2\sqrt{x}}=\frac{\cos\sqrt{x}}{2\sqrt{x}}.$$

Ответ

1. $$y’=10(2x+3)^4$$
2. $$y’=6\left(\frac13x-6\right)^{17}$$
3. $$y’=-2\sin 2x$$
4. $$y’=\sin 2x$$
5. $$y’=-\frac{3}{5\sin^2\frac{x}{5}}$$
6. $$y’=\frac{1}{\sqrt{2x+1}}$$
7. $$y’=-\frac{1}{3\sqrt[3]{(1-x)^2}}$$
8. $$y’=\frac{x}{\sqrt{x^2+1}}$$
9. $$y’=-\frac{4}{(4x+5)^2}$$
10. $$y’=-\frac{6x+24}{\left(\frac{x^2}{2}+4x-1\right)^7}$$
11. $$y’=\frac{\cos x}{2\sqrt{\sin x}}$$
12. $$y’=\frac{\cos\sqrt{x}}{2\sqrt{x}}$$