Question

Срочно, 40 балов!! АЛГЕБРА ​

Answer 1

Ответ:

$1)\ \ y=e^{-3x}\, tgx\\\\y'=-3e^{-3x}\, tgx+e^{-3x}\cdot \dfrac{1}{cos^2x}=e^{-3x}\cdot \Big(-3tgx+\dfrac{1}{cos^2x}\Big)$

$2)\ \ y=x^2\, lnx\ \ ,\ \ \ \ \ ODZ:\ x>0\ ,\\\\y'=2x\cdot lnx+x^2\cdot \dfrac{1}{x}=x\cdot (2\, lnx+1)=0\ \ \to \ \ \ x_1=0\ ,\\\\lnx=-\dfrac{1}{2}\ \ ,\ \ x=e^{-\frac{1}{2}}=\dfrac{1}{\sqrt{e}}\\\\znaki\ y'(x):\ \ \ (0)---(\dfrac{1}{\sqrt{e}})+++\\{}\qquad \qquad \ \ \qquad \qquad \searrow \ \ \qquad \quad \ \ \nearrow \\\\x_{min}=\dfrac{1}{\sqrt{e}}\ \ \ ,\ \ \ y_{min}=-\dfrac{1}{2e}$

$y(x)\ ybuvaet\ ,\ esli\ \ x\in \Big(\ 0;\dfrac{1}{\sqrt{e}}\ \Big]\\\\y(x)\ vozrastaet\ ,\ esli\ \ x\in \Big[\, \dfrac{1}{\sqrt{e}}\ ;\ +\infty \, \Big)$