КР1

Вариант 0

1. $y=3^x\tan (3x+8)$

$$y^{\prime }=3^x\ln 3\tan (3x+8)+\frac{3\cdot 3^x}{{\cos }^2(3x+8)}$$

2. $y=\sqrt[5]{{\sin }^2({\log }_{3}x)}$

$$y^{\prime }=\frac{2}{5}{\left(\sin ({\log }_{3}x)\right)}^{-\frac{3}{5}}\cdot \cos ({\log }_{3}x)\cdot \frac{1}{x\ln 3}=\frac{2\cos ({\log }_{3}x)}{5x\ln 3\sqrt[5]{{\sin }^3({\log }_{3}x)}}$$

3. $y=\arctan (x^3-2)-\frac{x}{\cos 5x}$

$$y^{\prime }=\frac{3x^2}{1+(x^3-2{)}^2}-\frac{\cos 5x+5x\sin 5x}{{\cos }^{2}5x}$$

4. $y=\frac{\arcsin \sqrt{x}}{x{e}^x}$

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

5. $y=(1+\cot 4x{)}^{\ln x}$

$$\begin{array}{c}{y}^{\prime }=(1+\cot 4x{)}^{\ln x}(\ln x\ln (1+\cot 4x){)}^{\prime }=\\ =(1+\cot 4x{)}^{\ln x}(\frac{\ln (1+\cot 4x))}{x}-\frac{4\ln x}{(1+\cot 4x){\sin }^{2}4x})\end{array}$$

6. $y={\arccos }^3\left(\frac{x}{x^2+1}\right)$

$$y^{\prime }=-3{\arccos }^2(\frac{x}{x^2+1})\cdot \frac{1-x^2}{(x^2+1{)}^2}\cdot \frac{1}{\sqrt{1-(\frac{x}{x^2+1}{)}^2}}$$

7. $\{\begin{array}{l}x={\log }_2(1+t^3)\\ y=t^4\end{array}{y}_{xx}^{\prime \prime }-?$

$$\begin{array}{c}\{\begin{array}{l}{x}_t^{\prime }=\frac{3t^2}{(1+t^3)\ln 2}\\ y_t^{\prime }=4t^3\end{array}\\ \{\begin{array}{l}{x}_t^{\prime }=\frac{3t^2}{(1+t^3)\ln 2}\\ y_x^{\prime }=\frac{y_t^{\prime }}{x_t^{\prime }}=\frac{4t^3\cdot (1+t^3)\ln 2}{3t^2}=\frac{4\ln 2}{3}(t+t^4)\end{array}\\ (y_x^{\prime }{)}_t^{\prime }=\frac{4\ln 2}{3}(1+4t^3)\\ y_{xx}^{\prime \prime }=\frac{(y_x^{\prime }{)}_t^{\prime }}{x_t^{\prime }}=\frac{\frac{4\ln 2}{3}(1+4t^3)}{\frac{3t^2}{(1+t^3)\ln 2}}=\frac{4(\ln 2{)}^2(1+4t^3)(1+t^3)}{9t^2}\end{array}$$

8.$x^2+y^2+6xy-4x+2y=0,M(0;0),y_x^{\prime }-?y_{xx}^{\prime \prime }-?$

$$\begin{array}{c}(x^2+y^2+6xy-4x+2y{)}_x^{\prime }=0\\ 2x+2y{y}_x^{\prime }+(6y+6x{y}_x^{\prime })-4+2y_x^{\prime }=0|:2\\ x+y{y}_x^{\prime }+3y+3x{y}_x^{\prime }-2+y_x^{\prime }=0\\ y_x^{\prime }=\frac{2-x-3y}{y+3x+1}\\ y^{\prime }{|}_{M(0,0)}=\frac{2-0-0}{0+0+1}=2\\ (x+y{y}_x^{\prime }+3y+3x{y}_x^{\prime }-2+y_x^{\prime }{)}_x^{\prime }=0\\ 1+(y_x^{\prime }{)}^2+y{y}_{xx}^{\prime \prime }+3y_x^{\prime }+3y_x^{\prime }+3x{y}_{xx}^{\prime \prime }+y_{xx}^{\prime \prime }=0\\ y_{xx}^{\prime \prime }=-\frac{1+(y_x^{\prime }{)}^2+6y_x^{\prime }}{y+1+3x}\\ y_{xx}^{\prime \prime }{|}_{M(0,0)}=-\frac{1+4+12}{0+1+0}=-17\end{array}$$

$9.\begin{array}{c}x=4\cos t;y=8\sin t;M_0(2\sqrt{3};4)\\ y_{\text{кас}}-?y_{\text{норм}}-?\{\begin{array}{l}x=4\cos t\\ y=8\sin t\end{array}\end{array}$

$$\begin{array}{c}\{\begin{array}{l}{x}_t^{\prime }=-4\sin t\\ y_t^{\prime }=8\cos t\end{array}\\ \\ \{\begin{array}{l}{x}_0=4\cos t_0\\ y_0=8\sin t_0\end{array}\{\begin{array}{l}2\sqrt{3}=4\cos t_0\\ 4=8\sin t_0\end{array}\\ \{\begin{array}{l}\cos t_0=\frac{\sqrt{3}}{2}\\ \sin t_0=0,5\end{array}\Rightarrow \{\begin{array}{l}{t}_0=\pm \frac{\pi }{6}+2\pi k,k\in \mathbb{Z}\\ [\begin{array}{rl}{t}_0& =\frac{\pi }{6}+2\pi k,k\in \mathbb{Z}\\ t_0& =\frac{5\pi }{6}+2\pi k,k\in \mathbb{Z}\end{array}\end{array}\\ \Rightarrow t_0=\frac{\pi }{6}+2\pi k,k\in \mathbb{Z}\\ \\ y_{\text{кас}}=y_x^{\prime }(t_0)(x-x_0)+y_0\\ y_{\text{кас}}=-2\mathrm{ctg}\frac{\pi }{6}(x-2\sqrt{3})+4\\ y_{\text{кас}}=-2\sqrt{3}x+16\\ \\ y_{\text{норм}}=-\frac{1}{y_x^{\prime }}(x-x_0)-y_0\\ y_{\text{норм}}=\frac{x}{2\sqrt{3}}+3\end{array}$$

Дополнительные задания

1. $y={\cos }^4\left(\sqrt[3]{2-x^2}\right)$

$$\begin{array}{c}{y}^{\prime }=4{\cos }^3\left(\sqrt[3]{2-x^2}\right)\cdot \left(-\sin \left(\sqrt[3]{2-x^2}\right)\right)\cdot \frac{1}{3\sqrt[3]{(2-x^2{)}^2}}\cdot (-2x)=\\ =\frac{8x{\cos }^3\left(\sqrt[3]{2-x^2}\right)\sin \left(\sqrt[3]{2-x^2}\right)}{3\sqrt[3]{(2-x^2{)}^2}}\end{array}$$

2. $y=x^5{\sin }^3\frac{1}{x^2}$

$$\begin{array}{c}{y}^{\prime }=5x^4{\sin }^3\frac{1}{x^2}+x^5\cdot 3{\sin }^2\frac{1}{x^2}\cos \frac{1}{x^2}\cdot \left(-2x^{-3}\right)=\\ =5x^4{\sin }^3\frac{1}{x^2}-6x^2{\sin }^2\frac{1}{x^2}\cos \frac{1}{x^2}\end{array}$$

3. $y=\mathrm{tg}\sqrt[4]{\ln \left(1+\frac{1}{2x}\right)}$

$$\begin{array}{c}{y}^{\prime }=\frac{1}{{\cos }^2\sqrt[4]{\ln \left(1+\frac{1}{2x}\right)}}\cdot \frac{1}{4}{\left(\ln \left(1+\frac{1}{2x}\right)\right)}^{-\frac{3}{4}}\cdot \frac{1}{1+\frac{1}{2x}}\cdot \left(-\frac{1}{2x^2}\right)=\\ =\frac{-1}{4x(2x+1)\sqrt[4]{{\ln }^3\left(1+\frac{1}{2x}\right)}{\cos }^2\sqrt[4]{\ln \left(1+\frac{1}{2x}\right)}}\end{array}$$

4. $y=\frac{\sqrt[7]{3x^2-x}}{{\cos }^{2}4x}$

$$\begin{array}{c}{y}^{\prime }=\frac{\frac{1}{7}(3x^2-x{)}^{-\frac{6}{7}}(6x-1){\cos }^{2}4x-\sqrt[7]{3x^2-x}\cdot 2\cos 4x(-\sin 4x)\cdot 4}{{\cos }^{4}4x}=\\ =\frac{\frac{(6x-1)\cos 4x}{7\sqrt[7]{(3x^2-x{)}^6}}+8\sqrt[7]{3x^2-x}\sin 4x}{{\cos }^{3}4x}\end{array}$$

5. $y={\left(\arcsin \frac{1}{3x}\right)}^{4x^2}$

$$\begin{array}{c}{y}^{\prime }={\left(\arcsin \frac{1}{3x}\right)}^{4x^2}\cdot \left({\left(4x^2\right)}^{\prime }\ln \left(\arcsin \frac{1}{3x}\right)+4x^2\cdot \frac{{\left(\arcsin \frac{1}{3x}\right)}^{\prime }}{\arcsin \frac{1}{3x}}\right)=\\ ={\left(\arcsin \frac{1}{3x}\right)}^{4x^2}\cdot \left(8x\ln \left(\arcsin \frac{1}{3x}\right)+4x^2\cdot \frac{1}{\arcsin \frac{1}{3x}}\cdot \frac{1}{\sqrt{1-\frac{1}{9x^2}}}\cdot \left(-\frac{1}{3x^2}\right)\right)=\\ ={\left(\arcsin \frac{1}{3x}\right)}^{4x^2}\cdot \left(8x\ln \left(\arcsin \frac{1}{3x}\right)-\frac{4x^2}{\arcsin \frac{1}{3x}}\cdot \frac{3x}{\sqrt{9x^2-1}}\cdot \frac{1}{3x^2}\right)=\\ ={\left(\arcsin \frac{1}{3x}\right)}^{4x^2}\cdot \left(8x\ln \left(\arcsin \frac{1}{3x}\right)-\frac{4x}{\sqrt{9x^2-1}\cdot \arcsin \frac{1}{3x}}\right)\end{array}$$

6. $y=\frac{\mathrm{ctg}5x}{1+{\cos }^{2}3x}$

$$\begin{array}{c}{y}^{\prime }=\frac{(\cot 5x{)}^{\prime }(1+{\cos }^{2}3x)-\cot 5x(1+{\cos }^{2}3x{)}^{\prime }}{(1+{\cos }^{2}3x{)}^2}=\\ =\frac{-\frac{5}{{\sin }^{2}5x}(1+{\cos }^{2}3x)-\cot 5x\cdot (2\cos 3x\cdot (-\sin 3x)\cdot 3)}{(1+{\cos }^{2}3x{)}^2}=\\ =\frac{-\frac{5(1+{\cos }^{2}3x)}{{\sin }^{2}5x}-\cot 5x\cdot (-3\sin 6x)}{(1+{\cos }^{2}3x{)}^2}=\\ =\frac{3\sin 6x\cot 5x-\frac{5(1+{\cos }^{2}3x)}{{\sin }^{2}5x}}{(1+{\cos }^{2}3x{)}^2}\end{array}$$

7. $y={\left(\sqrt{\tan 3x}\right)}^{{\sin }^{2}x}$

$$\begin{array}{c}{y}^{\prime }={\left(\sqrt{\tan 3x}\right)}^{{\sin }^{2}x}\cdot \left(({\sin }^{2}x{)}^{\prime }\ln \sqrt{\tan 3x}+{\sin }^{2}x\cdot \frac{(\sqrt{\tan 3x}{)}^{\prime }}{\sqrt{\tan 3x}}\right)=\\ ={\left(\sqrt{\tan 3x}\right)}^{{\sin }^{2}x}\cdot \left(2\sin x\cos x\ln \sqrt{\tan 3x}+{\sin }^{2}x\cdot \frac{1}{\sqrt{\tan 3x}}\cdot \frac{1}{2\sqrt{\tan 3x}}\cdot \frac{3}{{\cos }^{2}3x}\right)=\\ ={\left(\sqrt{\tan 3x}\right)}^{{\sin }^{2}x}\cdot \left(\sin 2x\ln \sqrt{\tan 3x}+\frac{3{\sin }^{2}x}{2\tan 3x{\cos }^{2}3x}\right)=\\ ={\left(\sqrt{\tan 3x}\right)}^{{\sin }^{2}x}\cdot \left(\sin 2x\ln \sqrt{\tan 3x}+\frac{3{\sin }^{2}x}{\sin 6x}\right)\end{array}$$

8. $y={\log }_3\sqrt{e^{3x}-5e^{-2x}}$

$$\begin{array}{c}y=\frac{1}{2}{\log }_3(e^{3x}-5e^{-2x})\\ y^{\prime }=\frac{1}{2}\cdot \frac{(e^{3x}-5e^{-2x}{)}^{\prime }}{(e^{3x}-5e^{-2x})\ln 3}\\ y^{\prime }=\frac{1}{2}\cdot \frac{3e^{3x}-5e^{-2x}\cdot (-2)}{(e^{3x}-5e^{-2x})\ln 3}\\ y^{\prime }=\frac{3e^{3x}+10e^{-2x}}{2\ln 3(e^{3x}-5e^{-2x})}\end{array}$$

9. $\{\begin{array}{l}x=3{\cos }^{2}3t\\ y=2{\sin }^{3}3t\end{array}{y}_{xx}^{\prime \prime }-?$

$$\begin{array}{c}\{\begin{array}{l}{x}_t^{\prime }=3\cdot 2\cos 3t\cdot (-\sin 3t)\cdot 3=-9\cdot (2\sin 3t\cos 3t)=-9\sin 6t\\ y_t^{\prime }=2\cdot 3{\sin }^{2}3t\cdot \cos 3t\cdot 3=18{\sin }^{2}3t\cos 3t\end{array}\\ y_x^{\prime }=\frac{y_t^{\prime }}{x_t^{\prime }}=\frac{18{\sin }^{2}3t\cos 3t}{-18\sin 3t\cos 3t}=-\sin 3t\\ (y_x^{\prime }{)}_t^{\prime }=(-\sin 3t{)}^{\prime }=-3\cos 3t\\ y_{xx}^{\prime \prime }=\frac{(y_x^{\prime }{)}_t^{\prime }}{x_t^{\prime }}=\frac{-3\cos 3t}{-9\sin 6t}=\frac{3\cos 3t}{9\cdot 2\sin 3t\cos 3t}=\frac{1}{6\sin 3t}\end{array}$$

10. $y_x^{\prime }-?y_{xx}^{\prime \prime }-?M_0(1,2);4x^2+3xy-y^2=6$

$$\begin{array}{c}(4x^2+3xy-y^2-6{)}_x^{\prime }=0\\ 8x+(3y+3x{y}^{\prime })-2y{y}^{\prime }=0\\ 3x{y}^{\prime }-2y{y}^{\prime }=-3y-8x\\ y^{\prime }(3x-2y)=-3y-8x\\ y^{\prime }=\frac{-3y-8x}{3x-2y}\\ \\ y^{\prime }{|}_{M_0}=\frac{-3(2)-8(1)}{3(1)-2(2)}=\frac{-6-8}{3-4}=\frac{-14}{-1}=14\\ \\ (8x+3y+3x{y}^{\prime }-2y{y}^{\prime }{)}_x^{\prime }=0\\ 8+3y^{\prime }+(3y^{\prime }+3x{y}^{\prime \prime })-2((y^{\prime }{)}^2+y{y}^{\prime \prime })=0\\ 8+6y^{\prime }+3x{y}^{\prime \prime }-2(y^{\prime }{)}^2-2y{y}^{\prime \prime }=0\\ y^{\prime \prime }(3x-2y)=2(y^{\prime }{)}^2-6y^{\prime }-8\\ y_{xx}^{\prime \prime }=\frac{2(y^{\prime }{)}^2-6y^{\prime }-8}{3x-2y}\\ y^{\prime \prime }{|}_{M_0}=\frac{2(14{)}^2-6(14)-8}{3(1)-2(2)}=\frac{2(196)-84-8}{-1}=\frac{392-92}{-1}=-300\end{array}$$

11. $y_{\text{кас}}-?y_1=\ln 2x3y_2-x=2$

$$\begin{array}{c}{y}_{кас}=y_2^{\prime }(x-x_0)+y(x_0)\\ k=y_2^{\prime }=\frac{1}{3}\\ \\ y^{\prime }=(\ln 2x{)}^{\prime }=\frac{1}{2x}\cdot (2x{)}^{\prime }=\frac{2}{2x}=\frac{1}{x}\\ \\ y_1^{\prime }(x_0)=k\\ \frac{1}{x_0}=\frac{1}{3}\Rightarrow x_0=3\\ \\ y_0=y(x_0)=\ln (2\cdot 3)=\ln 6\\ \\ y-y_0=k(x-x_0)\\ y-\ln 6=\frac{1}{3}(x-3)\\ y=\frac{1}{3}x-1+\ln 6\end{array}$$

12. $y_{кас}-?y_1=\mathrm{arcctg}3x5x-12y_2=1$.

$$\begin{array}{c}{y}_2^{\prime }=\frac{5}{12}\\ y_1^{\prime }=(\mathrm{arcctg}3x{)}^{\prime }=-\frac{1}{1+(3x{)}^2}\cdot (3x{)}^{\prime }=-\frac{3}{1+9x^2}\\ y_1^{\prime }=-\frac{1}{y_2^{\prime }}\\ -\frac{3}{1+9x_0^2}=-\frac{12}{5}\\ x_0=\pm \frac{1}{6}\\ y_{кас1}=-2,4x+0,4+\mathrm{arccot}0,5\\ \\ y_{кас2}=-2,4x-0,4+\pi -\mathrm{arccot}0,5\end{array}$$

13. $y_{\text{кас}}-?y_{\text{норм}}-?\frac{x^2}{16}-y^2=3;M_0(8,1)$

$$\begin{array}{c}\text{1. Находим производную неявной функции:}\\ {\left(\frac{x^2}{16}-y^2-3\right)}_x^{\prime }=0\\ \frac{2x}{16}-2y{y}_x^{\prime }=0\\ \frac{x}{8}-2y{y}_x^{\prime }=0\\ 2y{y}_x^{\prime }=\frac{x}{8}\\ y_x^{\prime }=\frac{x}{16y}\\ \\ \text{2. Вычисляем значение производной в точке }{M}_0(8,1):\\ y_x^{\prime }{|}_{M_0}=\frac{8}{16\cdot 1}=\frac{1}{2}=0,5\\ \\ \text{3. Составляем уравнение касательной:}\\ y_{\text{кас}}=y_x^{\prime }(M_0)(x-x_0)+y_0\\ y=0,5(x-8)+1\\ y=0,5x-4+1\\ y=0,5x-3\\ \\ \text{4. Составляем уравнение нормали:}\\ y_{\text{норм}}=-\frac{1}{y_x^{\prime }(M_0)}(x-x_0)+y_0\\ y=-\frac{1}{0,5}(x-8)+1\\ y=-2(x-8)+1\\ y=-2x+16+1\\ y=-2x+17\end{array}$$

14. $y_{\text{кас}}-?y_{\text{норм}}-?\{\begin{array}{l}x=4\sin t\\ y=3\cos t\end{array}{M}_0\left(2,\frac{3\sqrt{3}}{2}\right)$

$$\begin{array}{c}\text{1. Находим параметр }{t}_0:\\ \{\begin{array}{l}2=4\sin t_0\\ \frac{3\sqrt{3}}{2}=3\cos t_0\end{array}\Rightarrow \{\begin{array}{l}\sin t_0=0,5\\ \cos t_0=\frac{\sqrt{3}}{2}\end{array}\Rightarrow t_0=\frac{\pi }{6}+2\pi k,k\in \mathbb{Z}\\ \text{Возьмем }{t}_0=\frac{\pi }{6}\\ \\ \text{2. Находим производную }{y}_x^{\prime }\text{ в параметрическом виде:}\\ \{\begin{array}{l}{x}_t^{\prime }=4\cos t\\ y_t^{\prime }=-3\sin t\end{array}\\ y_x^{\prime }=\frac{y_t^{\prime }}{x_t^{\prime }}=\frac{-3\sin t}{4\cos t}=-\frac{3}{4}\mathrm{tg}t\\ \\ \text{3. Вычисляем значение производной в точке }{t}_0=\frac{\pi }{6}:\\ y_x^{\prime }\left(\frac{\pi }{6}\right)=-\frac{3}{4}\mathrm{tg}\frac{\pi }{6}=-\frac{3}{4}\cdot \frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{4}\\ \\ \text{4. Составляем уравнение касательной:}\\ y_{\text{кас}}=y_x^{\prime }(t_0)(x-x_0)+y_0\\ y=-\frac{\sqrt{3}}{4}(x-2)+\frac{3\sqrt{3}}{2}\\ y=-\frac{\sqrt{3}}{4}x+\frac{\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}\\ y=-\frac{\sqrt{3}}{4}x+\frac{4\sqrt{3}}{2}\\ y=-\frac{\sqrt{3}}{4}x+2\sqrt{3}\\ \\ \text{5. Составляем уравнение нормали:}\\ y_{\text{норм}}=-\frac{1}{y_x^{\prime }(t_0)}(x-x_0)+y_0\\ y=-\frac{1}{-\frac{\sqrt{3}}{4}}(x-2)+\frac{3\sqrt{3}}{2}\\ y=\frac{4}{\sqrt{3}}(x-2)+\frac{3\sqrt{3}}{2}\\ y=\frac{4\sqrt{3}}{3}x-\frac{8\sqrt{3}}{3}+\frac{3\sqrt{3}}{2}\\ y=\frac{4\sqrt{3}}{3}x+\frac{-16\sqrt{3}+9\sqrt{3}}{6}\\ y=\frac{4\sqrt{3}}{3}x-\frac{7\sqrt{3}}{6}\end{array}$$