mzd418835233
2011-04-03, 17:15
这个式子求A对g的一阶导数 ，
syms v g f
y=(((2.*v.*g).^2+(g.^2-f.^2).^2)./((2.*v.*g).^2.*(g.^2-1+v.*g.^2).^2+(v.*f.^2.*g.^2-(g.^2-1).*(g.^2-f.^2)).^2)).^0.5;
diff(y,g)
小弟的结果
ans =
1/2/((4*v^2*g^2+(g^2-f^2)^2)/(4*v^2*g^2*(g^2-1+v*g^2)^2+(v*f^2*g^2-(g^2-1)*(g^2-f^2))^2))^(1/2)*((8*v^2*g+4*(g^2-f^2)*g)/(4*v^2*g^2*(g^2-1+v*g^2)^2+(v*f^2*g^2-(g^2-1)*(g^2-f^2))^2)-(4*v^2*g^2+(g^2-f^2)^2)/(4*v^2*g^2*(g^2-1+v*g^2)^2+(v*f^2*g^2-(g^2-1)*(g^2-f^2))^2)^2*(8*v^2*g*(g^2-1+v*g^2)^2+8*v^2*g^2*(g^2-1+v*g^2)*(2*g+2*v*g)+2*(v*f^2*g^2-(g^2-1)*(g^2-f^2))*(2*v*f^2*g-2*(g^2-f^2)*g-2*(g^2-1)*g)))
能弄成我们常用的分式吗？ 求高手啊！！！！
syms v g f
y=(((2.*v.*g).^2+(g.^2-f.^2).^2)./((2.*v.*g).^2.*(g.^2-1+v.*g.^2).^2+(v.*f.^2.*g.^2-(g.^2-1).*(g.^2-f.^2)).^2)).^0.5;
diff(y,g)
小弟的结果
ans =
1/2/((4*v^2*g^2+(g^2-f^2)^2)/(4*v^2*g^2*(g^2-1+v*g^2)^2+(v*f^2*g^2-(g^2-1)*(g^2-f^2))^2))^(1/2)*((8*v^2*g+4*(g^2-f^2)*g)/(4*v^2*g^2*(g^2-1+v*g^2)^2+(v*f^2*g^2-(g^2-1)*(g^2-f^2))^2)-(4*v^2*g^2+(g^2-f^2)^2)/(4*v^2*g^2*(g^2-1+v*g^2)^2+(v*f^2*g^2-(g^2-1)*(g^2-f^2))^2)^2*(8*v^2*g*(g^2-1+v*g^2)^2+8*v^2*g^2*(g^2-1+v*g^2)*(2*g+2*v*g)+2*(v*f^2*g^2-(g^2-1)*(g^2-f^2))*(2*v*f^2*g-2*(g^2-f^2)*g-2*(g^2-1)*g)))
能弄成我们常用的分式吗？ 求高手啊！！！！