汇编子程序调用:d[(h/s)^2+1]^0.5/ds=?(h为常数)
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是不是=-s^(-2)
d[(h/s)^2+1]^0.5/ds=0.5*[(h/s)^2+1]^(-0.5)*2(h/s)*(-h/s^2)=-[(h/s)^2+1]^(-0.5)*(h^2/s^3)
d[(h/s)^2+1]^0.5/ds=0.5*[(h/s)^2+1]^(-0.5)*2(h/s)*(-h/s^2)=-[(h/s)^2+1]^(-0.5)*(h^2/s^3)