Question

помогите найти частные производные первого и второго порядка

Answer 1

$z=\ln(y+\cos xy)\\z'_x=\frac1{y+\cos xy}\cdot(y+\cos xy)'_x=-\frac{\sin xy\cdot(xy)'_x}{y+\cos xy}=-\frac{y\sin xy}{y+\cos xy}\\z_y'=\frac1{y+\cos xy}\cdot(y+\cos xy)'_y=\frac{1-\sin xy\cdot(xy)'_y}{y+\cos xy}=\frac{1-x\sin xy}{y+\cos xy}\\\\z''_x=-\frac{y\cos xy\cdot(xy)'_x\cdot(y+\cos xy)-y\sin xy\cdot(-\sin xy)\cdot(xy)'x}{(y+\cos xy)^2}=\\=-\frac{y^2\cos xy(y+\cos xy)+y^2\sin^2xy}{(y+\cos xy)^2}=-\frac{y^3\cos xy+y^2\cos^2xy+y^2\sin^2xy}{(y+\cos xy)^2}=\\=-\frac{y^2(\cos xy+1)}{(y+\cos xy)^2}$
$z_y''=\frac{-x\cos xy\cdot(xy)'_y\cdot(y+\cos xy)-(1-x\sin xy)(1-\sin xy\cdot(xy)'_y)}{(y+\cos xy)^2}=\\=\frac{-x^2\cos xy(y+\cos xy)-(1-x\sin xy)(1-x\sin xy)}{(y+cos xy)^2}=\\=\frac{-x^2y\cos xy-x^2\cos^2xy-1+2x\sin xy-x^2\sin^2xy}{(y+\cos xy)^2}=\\=\frac{2x\sin xy-x^2y\cos xy-1-x^2(\sin^2xy+\cos^2xy)}{(y+\cos xy)^2}=\frac{2x\sin xy-x^2y\cos xy-1-x^2}{(y+\cos xy)^2}$