Упр.200 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1. $$3x^2-4\sqrt[3]{x}+2e^{x/3}$$
2. $$2x^3+3\sqrt{x}-\cos 2x$$
3. $$\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}+\ln x^3$$
4. $$2x^8-3\tg 3x-\frac{1}{3}\sin 3x$$
5. $$8x^{3/4}+7x^{1/7}-\cos 4x$$
6. $$\frac{1}{5}\ctg x-5x^{4/5}-\frac{1}{4}e^{2x}$$

1. $$y=3x^2-4\sqrt[3]{x}+2e^{x/3}$$

   $$y’=(3x^2)’-(4x^{1/3})’+(2e^{x/3})’$$
   $$=6x-\frac{4}{3}x^{-2/3}+\frac{2}{3}e^{x/3}$$
   $$=6x-\frac{4}{3\sqrt[3]{x^2}}+\frac{2}{3}e^{x/3}.$$

2. $$y=2x^3+3\sqrt{x}-\cos 2x$$

   $$y’=(2x^3)’+(3\sqrt{x})’-(\cos 2x)’$$
   $$=6x^2+\frac{3}{2\sqrt{x}}+2\sin 2x.$$

3. $$y=\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}+\ln x^3$$

   $$y’=(x^{1/3})’+(x^{-1/3})’+(\ln x^3)’$$
   $$=\frac{1}{3}x^{-2/3}-\frac{1}{3}x^{-4/3}+\frac{3}{x}$$
   $$=\frac{1}{3\sqrt[3]{x^2}}-\frac{1}{3x\sqrt[3]{x}}+\frac{3}{x}.$$

4. $$y=2x^8-3\tg 3x-\frac{1}{3}\sin 3x$$

   $$y’=(2x^8)’-(3\tg 3x)’-\left(\frac{1}{3}\sin 3x\right)’$$
   $$=16x^7-\frac{9}{\cos^2 3x}-\cos 3x.$$

5. $$y=8x^{3/4}+7x^{1/7}-\cos 4x$$

   $$y’=(8x^{3/4})’+(7x^{1/7})’-(\cos 4x)’$$
   $$=6x^{-1/4}+x^{-6/7}+4\sin 4x$$
   $$=\frac{6}{\sqrt[4]{x}}+\frac{1}{\sqrt[7]{x^6}}+4\sin 4x.$$

6. $$y=\frac{1}{5}\ctg x-5x^{4/5}-\frac{1}{4}e^{2x}$$

   $$y’=\frac{1}{5}(\ctg x)’-5(x^{4/5})’-\frac{1}{4}(e^{2x})’$$
   $$=-\frac{1}{5\sin^2 x}-4x^{-1/5}-\frac{1}{2}e^{2x}$$
   $$=-\frac{1}{5\sin^2 x}-\frac{4}{\sqrt[5]{x}}-\frac{1}{2}e^{2x}.$$

Ответ

1. $$y’=6x-\frac{4}{3\sqrt[3]{x^2}}+\frac{2}{3}e^{x/3}$$
2. $$y’=6x^2+\frac{3}{2\sqrt{x}}+2\sin 2x$$
3. $$y’=\frac{1}{3\sqrt[3]{x^2}}-\frac{1}{3x\sqrt[3]{x}}+\frac{3}{x}$$
4. $$y’=16x^7-\frac{9}{\cos^2 3x}-\cos 3x$$
5. $$y’=\frac{6}{\sqrt[4]{x}}+\frac{1}{\sqrt[7]{x^6}}+4\sin 4x$$
6. $$y’=-\frac{1}{5\sin^2 x}-\frac{4}{\sqrt[5]{x}}-\frac{1}{2}e^{2x}$$