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

1) y=1/x^9-3/x^3; 5) y=cos(3x)/(x-1);
2) y=xv(2x+1); 6) y=(vx-1)/(vx+1);
3) y=sin(x)cos(2x); 7) y=(x+1)^3 (x-2)^4;
4) y=tg(x)sin(2x+5); 8) y=v(x^2+1)/x.

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. $$y=x\sqrt{2x+1}$$

   $$y’=x’\sqrt{2x+1}+x\left(\sqrt{2x+1}\right)’$$

   $$y’=\sqrt{2x+1}+x\cdot \frac{2}{2\sqrt{2x+1}}=\sqrt{2x+1}+\frac{x}{\sqrt{2x+1}}$$

   $$y’=\frac{2x+1+x}{\sqrt{2x+1}}=\frac{3x+1}{\sqrt{2x+1}}$$

3. $$y=\sin x\cdot \cos 2x$$

   $$y’=(\sin x)’\cos 2x+\sin x\cdot (\cos 2x)’$$

   $$y’=\cos x\cos 2x+\sin x\cdot (-2\sin 2x)$$

   $$y’=\cos x\cos 2x-2\sin x\sin 2x$$

4. $$y=\tg x\cdot \sin(2x+5)$$

   $$y’=(\tg x)’\sin(2x+5)+\tg x\cdot (\sin(2x+5))’$$

   $$y’=\frac{1}{\cos^2 x}\sin(2x+5)+\tg x\cdot 2\cos(2x+5)$$

   $$y’=\frac{\sin(2x+5)}{\cos^2 x}+2\tg x\cos(2x+5)$$

5. $$y=\frac{\cos 3x}{x-1}$$

   $$y’=\frac{(\cos 3x)'(x-1)-\cos 3x\cdot (x-1)’}{(x-1)^2}$$

   $$y’=\frac{3(-\sin 3x)(x-1)-\cos 3x}{(x-1)^2}$$

   $$y’=\frac{3(1-x)\sin 3x-\cos 3x}{(x-1)^2}$$

6. $$y=\frac{\sqrt{x}-1}{\sqrt{x}+1}$$

   $$y’=\frac{(\sqrt{x}-1)'(\sqrt{x}+1)-(\sqrt{x}-1)(\sqrt{x}+1)’}{(\sqrt{x}+1)^2}$$

   $$y’=\frac{\frac{1}{2\sqrt{x}}(\sqrt{x}+1)-(\sqrt{x}-1)\frac{1}{2\sqrt{x}}}{(\sqrt{x}+1)^2}$$

   $$y’=\frac{\sqrt{x}+1-\sqrt{x}+1}{2\sqrt{x}(\sqrt{x}+1)^2}=\frac{1}{\sqrt{x}(\sqrt{x}+1)^2}$$

7. $$y=(x+1)^3(x-2)^4$$

   $$y’=3(x+1)^2(x-2)^4+(x+1)^3\cdot 4(x-2)^3$$

   $$y’=(x+1)^2(x-2)^3\bigl(3(x-2)+4(x+1)\bigr)$$

   $$y’=(x+1)^2(x-2)^3(7x-2)$$

8. $$y=\frac{\sqrt{x^2+1}}{x}$$

   $$y’=\frac{(\sqrt{x^2+1})’\cdot x-\sqrt{x^2+1}\cdot (x)’}{x^2}$$

   $$y’=\frac{\frac{2x}{2\sqrt{x^2+1}}\cdot x-\sqrt{x^2+1}}{x^2}$$

   $$y’=\frac{\frac{x^2}{\sqrt{x^2+1}}-\sqrt{x^2+1}}{x^2}=-\frac{1}{x^2\sqrt{x^2+1}}$$

Ответ

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