Question

Найдите частные производные первого и второго порядка

Answer 1

$z(x,y)=cos^3(-5x+3\sqrt{y})\\\\\\z'_{x}=3\, cos^2(-5x+3\sqrt{y})\cdot (-sin(-5x+3\sqrt{y}))\cdot (-5)=\\\\=15\cdot cos^2(-5x+3\sqrt{y})\cdot sin(-5x+3\sqrt{y})=\\\\=\frac{15}{2}\cdot cos(-5x+3\sqrt{y})\cdot sin(-10x+6\sqrt{y})\\\\\\z'_{y}=3\, cos^2(-5x+3\sqrt{y})\cdot (-sin(-5x+3\sqrt{y}))\cdot 3\cdot \frac{1}{2\sqrt{y}}=\\\\=-\frac{9}{2\sqrt{y}}\cdot cos^2(-5x+3\sqrt{y})\cdot sin(-5x+3\sqrt{y})=\\\\=-\frac{9}{4\sqrt{y}}\cdot cos(-5x+3\sqrt{y})\cdot sin(-10x+6\sqrt{y})$

$z''_{xx}=\frac{15}{2}\cdot \Big (-sin(-5x+3\sqrt{y})\cdot (-5)\cdot sin(-10x+6\sqrt{y})+\\\\+cos(-5x+3\sqrt{y})\cdot cos(-10x+6\sqrt{y})\cdot (-10)\Big )=\\\\=\frac{15}{2}\cdot \Big (5\, sin(-5x+3\sqrt{y})\cdot sin(-10x+6\sqrt{y})-\\\\-10\, cos(-5x+3\sqrt{y})\cdot cos(-10x+6\sqrt{y})\Big )$

$z''_{xy}=\frac{15}{2}\cdot \Big (-sin(-5x+3\sqrt{y})\cdot \frac{3}{2\sqrt{y}}\cdot sin(-10x+6\sqrt{y})+\\\\+cos(-5x+3\sqrt{y})\cdot cos(-10x+6\sqrt{y})\cdot \frac{6}{2\sqrt{y}}\Big )$

$z''_{yy}=-\frac{9}{4}\cdot \Big (-\frac{1}{2\sqrt{y^3}}\cdot cos(-5x+3\sqrt{y})\cdot sin(-10x+6\sqrt{y})+\\\\+\frac{1}{\sqrt{y}}\cdot (-sin(-5x+3\sqrt{y}))\cdot \frac{3}{2\sqrt{y}}\cdot sin(-10x+6\sqrt{y})+\\\\+\frac{1}{\sqrt{y}}\cdot cos(-5x+3\sqrt{y})\cdot cos(-10x+6\sqrt{y})\cdot \frac{6}{2\sqrt{y}}\Big )$