利用导数的定义求函数的导数f(x)=三次根号下x?
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f(x)=x^(1/3)
f'(x)=lim [f(x+△x)-f(x)]/△x=lim [(x+△x)^(1/3)-x^(1/3)]/△x
△x->0 △x->0
=lim [(x+△x)^(1/3)-x^(1/3)]*[(x+△x)^(2/3)+(x+△x)^(1/3)*x^(1/3)+x^(2/3)]/{[(x+△x)^(2/3)+(x+△x)^(1/3)*x^(1/3)+x^(2/3)]*△x}
△x->0
=lim (x+△x-x)/{[(x+△x)^(2/3)+(x+△x)^(1/3)*x^(1/3)+x^(2/3)]*△x}
△x->0
=lim 1/[(x+△x)^(2/3)+(x+△x)^(1/3)*x^(1/3)+x^(2/3)]
△x->0
=1/3*x^(-2/3),1,
f'(x)=lim [f(x+△x)-f(x)]/△x=lim [(x+△x)^(1/3)-x^(1/3)]/△x
△x->0 △x->0
=lim [(x+△x)^(1/3)-x^(1/3)]*[(x+△x)^(2/3)+(x+△x)^(1/3)*x^(1/3)+x^(2/3)]/{[(x+△x)^(2/3)+(x+△x)^(1/3)*x^(1/3)+x^(2/3)]*△x}
△x->0
=lim (x+△x-x)/{[(x+△x)^(2/3)+(x+△x)^(1/3)*x^(1/3)+x^(2/3)]*△x}
△x->0
=lim 1/[(x+△x)^(2/3)+(x+△x)^(1/3)*x^(1/3)+x^(2/3)]
△x->0
=1/3*x^(-2/3),1,
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