Question

高数求助

第一题...
速度速度...

Answer 1

因为f[g(x)]'=f'[g(x)]*g'(x)=(-2/x^3)*f'[g(x)]=-1/2x
所以f'[g(x)]=x^2/4
f'(1/x^2)=x^2/4
令t=1/x^2
x^2=1/t
f'(t)=1/4t
f'(x)=1/4x
f'(1/2)=1/2
选C

Answer 2

g(x)=1/x^2
f(g(x))=f(1/x^2)
f(g(x))′=-1/(2x)
则，f(g(x))=f(1/x^2)=（-1/2）lnx
令1/x^2=t
x=1/√t
f(t)=(-1/2)*(-1/2)lnt=(1/4)lnt
f(x)=(1/4)lnx
f(x)'=1/(4x)
f(1/2)'=1/2

Answer 3

1。[f(g(x))]'=f'(g(x))*g'(x)=f'(1/x^2)*(-2/x^3)=-1/(2x)
f'(1/x^2)=x^2/4
即：f'(x)=1/(4x)
f'(1/2)=1/(4*1/2)=1/2，选C
2。f'(x)=cosx
(f(f(x)))'=f'(f(x))*f'(x)
=f'(sinx)*cosx
=cos(sinx)*cosx
选A