Question

设函数f(x-1/x)=lnx,求f(x)的导数。详细解答

Answer 1

f(x-1/x)=lnx
令u=x-1/x
ux=x²-1
x²-ux-1=0,Δ=u²-4*(-1)=u²+4
x=[-(-u)±√(u²+4)]/2
=1/2*[u²±√(u²+4)]
=1/2*[u²+√(u²+4)]>0或1/2*[u²+√(u²+4)]<0(真数少于0无意义.故舍去)

∴f(u)=ln{1/2*[u²+√(u²+4)]}
∴f(x)=ln{1/2*[x²+√(x²+4)]}
f'(x)={1/2*[x²+√(x²+4)]}'/{1/2*[x²+√(x²+4)]}
=[2x+(x²+4)'/2√(x²+4)]/[x²+√(x²+4)]
=[2x+x/√(x²+4)]/[x²+√(x²+4)]
=[2x√(x²+4)+x]/{[x²+√(x²+4)]√(x²+4)}
=x[2√(x²+4)+1]/{[x²+√(x²+4)]√(x²+4)}

Answer 2

我知道答案是1/((x^2+4)^(1/2)) 求过程