Question

Найти dz/du и dz/dv, если z=y^2lnx, где x=2u-3v, y=u/v

Answer 1

Частные производные...
$z=y^2*lnx=(frac{u}{v})^2*ln(2u-3v)\frac{dz}{du}=frac{1}{v^2}*2u*ln(2u-3v)+frac{1}{2u-3v}*2*(frac{u}{v})^2=frac{2u*ln(2u-3v)}{v^2}+\+frac{2*u^2}{v^2(2u-3v)}=frac{2u(2u-3v)ln(2u-3v)+2u^2}{v^2(2u-3v)}=frac{2u((2u-3v)ln(2u-3v)+u)}{v^2(2u-3v)}\\frac{dz}{dv}=u^2*(-frac{1}{v})*ln(2u-3v)+frac{1}{2u-3v}*(-3)*(frac{u}{v})^2=\=-frac{u^2*ln(2u-3v)}{v}-frac{3u^2}{v^2(2u-3v)}=frac{-u^2v*ln(2u-3v)-3u^2}{v^2(2u-3v)}=\=frac{-u^2(v*ln(2u-3v)+3)}{v^2(2u-3v)}$