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Question Number 89178 by necxxx last updated on 15/Apr/20
Ifz(z2+3x)+3y=0provethat∂2z∂x2+∂2z∂y2=2z(x−1)(z2+x)3pleasehelp.
Commented by niroj last updated on 16/Apr/20
Ifz(z2+3x)+3y=0Provethat∂2z∂x2+∂2z∂y2=2z(x−1)(z2+x)3D.w.r.toxz3+3xz+3y=03z2.dzdx+3z+3xdzdx=03dzdx(z2+x)=−3zdzdx=−zz2+x=d2zdx2=−dzdx(z2+x)+z(2z.dzdx+1)(z2+x)2=−dzdx(z2+x)+2z2dzdx+z(z2+x)2=dzdx(2z2−z2−x)+z(z2+x)2=−zz2+x(z2−x)+z(z2+x)2=xz−z3+z3+zx(z2+x)3=2zx(z2+x)3Again,D.w.r.to.yz3+3xz+3y=03z2dzdy+3xdzdy+3=03dzdy(z2+x)=−3dzdy=−1.(z2+x)−1d2zdy2=−1.(−1)(z2+x)−2.2zdzdy=2z(z2+x)2.−1(z2+x)=−2z(z2+x)3now,∂2z∂x2+∂2z∂2y=2xz(z2+x)3+−2z(z2+x)3=2xz−2z(z2+x)3=2z(x−1)(z2+x)3hence∴∂2z∂x2+∂2z∂y2=2z(x−1)(z2+x)3Proved//.
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