Упр.36.19 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1. Найдите производную функции:
   1) $$y=\frac{1}{x^9}-\frac{3}{x^3};$$
   5) $$y=\frac{\cos(3x)}{x-1};$$
   2) $$y=x\sqrt{2x+1};$$
   6) $$y=\frac{\sqrt{x}-1}{\sqrt{x}+1};$$
   3) $$y=\sin(x)\cos(2x);$$
   7) $$y=(x+1)^3(x-2)^4;$$
   4) $$y=\tan(x)\sin(2x+5);$$
   8) $$y=\frac{\sqrt{x^2+1}}{x}.$$

Найдём производные по правилам дифференцирования.

1. $$y=\frac{1}{x^9}-\frac{3}{x^3}=x^{-9}-3x^{-3}$$
   $$y’=-9x^{-10}+9x^{-4}=\frac{9x^6-9}{x^{10}}=\frac{9(x^6-1)}{x^{10}}.$$
2. $$y=x\sqrt{2x+1}$$
   $$y’=1\cdot \sqrt{2x+1}+x\cdot \frac{2}{2\sqrt{2x+1}}$$
   $$y’=\sqrt{2x+1}+\frac{x}{\sqrt{2x+1}}=\frac{3x+1}{\sqrt{2x+1}}.$$
3. $$y=\sin x\cos 2x$$
   $$y’=(\sin x)’\cos 2x+\sin x\cdot (\cos 2x)’$$
   $$y’=\cos x\cos 2x-2\sin x\sin 2x.$$
4. $$y=\tg x\cdot \sin(2x+5)$$
   $$y’=(\tg x)’\sin(2x+5)+\tg x\cdot (\sin(2x+5))’$$
   $$y’=\frac{\sin(2x+5)}{\cos^2 x}+2\tg x\cos(2x+5).$$
5. $$y=\frac{\cos 3x}{x-1}$$
   $$y’=\frac{(\cos 3x)'(x-1)-\cos 3x\cdot (x-1)’}{(x-1)^2}$$
   $$y’=\frac{-3\sin 3x\,(x-1)-\cos 3x}{(x-1)^2}=\frac{3(1-x)\sin 3x-\cos 3x}{(x-1)^2}.$$
6. $$y=\frac{\sqrt{x}-1}{\sqrt{x}+1}$$
   $$y’=\frac{(\sqrt{x}-1)'(\sqrt{x}+1)-(\sqrt{x}-1)(\sqrt{x}+1)’}{(\sqrt{x}+1)^2}$$
   $$y’=\frac{\frac{1}{2\sqrt{x}}(\sqrt{x}+1)-(\sqrt{x}-1)\frac{1}{2\sqrt{x}}}{(\sqrt{x}+1)^2}$$
   $$y’=\frac{1}{\sqrt{x}(\sqrt{x}+1)^2}.$$
7. $$y=(x+1)^3(x-2)^4$$
   $$y’=3(x+1)^2(x-2)^4+4(x+1)^3(x-2)^3$$
   $$y’=(x+1)^2(x-2)^3\bigl(3(x-2)+4(x+1)\bigr)$$
   $$y’=(x+1)^2(x-2)^3(7x-2).$$
8. $$y=\frac{\sqrt{x^2+1}}{x}$$
   $$y’=\frac{(\sqrt{x^2+1})’\cdot x-\sqrt{x^2+1}\cdot (x)’}{x^2}$$
   $$y’=\frac{\frac{2x}{2\sqrt{x^2+1}}\cdot x-\sqrt{x^2+1}}{x^2}$$
   $$y’=\frac{x^2-(x^2+1)}{x^2\sqrt{x^2+1}}=-\frac{1}{x^2\sqrt{x^2+1}}.$$

Ответ:

$$1)\ y’=\frac{9(x^6-1)}{x^{10}};\quad 2)\ y’=\frac{3x+1}{\sqrt{2x+1}};\quad 3)\ y’=\cos x\cos 2x-2\sin x\sin 2x;$$
$$4)\ y’=\frac{\sin(2x+5)}{\cos^2 x}+2\tg x\cos(2x+5);\quad 5)\ y’=\frac{3(1-x)\sin 3x-\cos 3x}{(x-1)^2};$$
$$6)\ y’=\frac{1}{\sqrt{x}(\sqrt{x}+1)^2};\quad 7)\ y’=(x+1)^2(x-2)^3(7x-2);\quad 8)\ y’=-\frac{1}{x^2\sqrt{x^2+1}}.$$