解:1、f'(x)=2x所以,f(f'(x))=2(2x)=4x,f'(f'(x))=4
2、d(x^2)/d(2x)=(2xdx)/(2dx)=x
d(x^2)/d(1/2x)=(2xdx)/{[-1/(2x^2)]dx}=-4x^3
f’(x)=(x^2)’=2x f(f’(x))=f(2x)=(2x)^2=4x^2 f’(f’(x))=f’(4x^2)=8x
1 f’(x)=(x^2)’=2x;
f(f’(x))=f(2x)=(2x)^2=4x^2;
f’(f’(x))=f’(4x^2)=8x
2 x^2=1/4*(2x)^2;d (x^2)/ d(2x) = d (1/4*(2x)^2)/ d(2x) = d (1/4*t^2)/ dt=2t=x;
x^2=4*(x/2)^2; d (x^2)/ d(x/2)= d(4*(x/2)^2)/ d(x/2)= d (4*t2)/ d t=8t=4x;
f'(x)=2x f'(f'(x))=f'(2x)=4x d(x^2)/d(2x)=2xdx/2dx=x d(x^2)/d(x/2)=2xdx/[(1/2)dx]=4x