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Question Number 207175 by MATHEMATICSAM last updated on 08/May/24
Ifxm.yn=(x+y)m+nthend2ydx2=?
Answered by som(math1967) last updated on 08/May/24
lnxm.yn=ln(x+y)m+nmlnx+nlny=(m+n)ln(x+y)mx+ny×dydx=m+nx+y×(1+dydx)(ny−m+nx+y)dydx=m+nx+y−mxnx−myy(x+y)×dydx=nx−myx(x+y)dydx=yxd2ydx2=dydx×x−yx2d2ydx2=yx×x−yx2=0
Answered by mr W last updated on 08/May/24
lety=txxm+ntn=(1+t)xm+n⇒tn=1+t⇒t=constantdydx=td2ydx2=0
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