df/dv= f `_(v)=f `_(x)*x`_(v)+f `_(y)*y`_(v)
Находим
f `_(x)= siny*(sqrt(x))`_(x)+(ln(1/y))`_(x)=siny/(2sqrt(x)) + 0
f `_(y)=sqrt(x)*(siny)`_(y)+(ln(1/y))`_(y)=sqrt(x)*(siny)`_(y)-(lny)`_(y)=
=sqrt(x)*(cosy) - (1/y)
x`_(u)= (cosv)`_(u)=0
y `_(u)=(v^3/u)`_(u)= v^3*(1/u)`_(u)=-v^3/u2
x`_(v)=(cosv)`_(v)= - sinv
y`_(v)=(v^3/u)`_(v)= (1/u)*(v^3)`_(v)=3v^2/u
Подставляем и получаем ответ:
df/du=0*siny/(2sqrt(x)) -(-v^3/u2)*( sqrt(x)*(cosy) - (1/y))
df/dv=-sinv*( siny)/(2sqrt(x)) +(3v^2/u)*( sqrt(x)*(cosy) - (1/y))