Question

求二元偏导数Z=(X^2+Y^2)^COSXY

Answer 1

设u=x^2+y^2, v=cos(xy), z=u^v

əu/əx=2x, əu/əy=2y

əv/əx=-ysin(xy), əv/əy=-xsin(xy)

əz/əu=v*u^(v-1), əz/əv=u^v*lnu

əz/əx=əz/əu*əu/əx+əz/əv*əv/əx

=v*u^(v-1)*2x+u^v*lnu*[-ysin(xy)]

=2x*cos(xy)*(x^2+y^2)^[cos(xy)-1]-ysin(xy)*(x^2+y^2)^[cos(xy)]*ln(x^2+y^2)

əz/əy=əz/əu*əu/əy+əz/əv*əv/əy

=v*u^(v-1)*2y+u^v*lnu*[-xsin(xy)]

=2y*cos(xy)*(x^2+y^2)^[cos(xy)-1]-xsin(xy)*(x^2+y^2)^[cos(xy)]*ln(x^2+y^2)