Question

Нужно найти вторую производную у функций задачи №8.280,8.281 и 8.284. помогите решить любую какую сможете

Answer 1

Ответ:

$\displaystyle 280.\ \ z=xy\, ln\frac{x}{y}\\\\\\z'_{x}=y\, ln\frac{x}{y}+xy\cdot \frac{y}{x}\cdot \frac{1}{y}=y\, ln\frac{x}{y}+y=y\cdot (ln\frac{x}{y}+1)\\\\\\z'_{y}=x\, ln\frac{x}{y}+xy\cdot \frac{y}{x}\cdot \frac{-x}{y^2}=x\, ln\frac{x}{y}-x=x\cdot (ln\frac{x}{y} -1)\\\\\\z''_{xx}=y\cdot \frac{y}{x}\cdot \frac{1}{y}=\frac{y}{x}$

$\displaystyle z''_{yy}=x\cdot \dfrac{y}{x}\cdot \frac{-x}{y^2}=-\frac{x}{y}\\\\\\z''_{xy}=1\cdot (ln\frac{x}{y}+1)+y\cdot (\frac{y}{x}\cdot \frac{-x}{y^2})=ln\frac{x}{y}+1-1=ln\frac{x}{y}\\\\\\z''_{yx}=1\cdot (ln\frac{x}{y}-1)+x\cdot (\frac{y}{x}\cdot \frac{1}{y})=ln\frac{x}{y}-1+1=ln\frac{x}{y}=z''_{xy}$

$281.\ \ z=3x^2+2xy^2-4xy+x^2y-y^3\\\\z'_{x}=6x+2y^2-4y+2xy\\\\z'_{y}=4xy-4x+x^2-3y^2\\\\z''_{xx}=6+2y\\\\z''_{yy}=4x-6y\\\\z''_{xy}=4y+2x\\\\z''_{yx}=4y+2x=z''_{yx}$

$\displaystyle 284.\ \ \ z=x^2\cdot sin\sqrt{y}\\\\z'_{x}=2x\cdot sin\sqrt{y}\\\\z'_{y}=x^2\cdot cos\sqrt{y}\cdot \frac{1}{2\sqrt{y}}\\\\z''_{xx}=2\cdot sin\sqrt{y}\\\\z''_{yy}=x^2\cdot \Big(-sin\sqrt{y}\cdot \frac{1}{2\sqrt{y}}\cdot \frac{1}{2\sqrt{y}}+cos\sqrt{y}\cdot \frac{-1}{4y\sqrt{y}}\ \Big)=\\\\\\=-\frac{x^2}{4y}\cdot \Big(\, sin\sqrt{y}+\frac{cos\sqrt{y}}{\sqrt{y}}\ \Big)\\\\\\z''_{xy}=2x\cdot cos\sqrt{y}\cdot\frac{1}{2\sqrt{y}}=z''_{yx}$