设函数f(x)在(0,+∞)内可微,其反函数为g(x),且∫[上下限(1,f(x))]g(t)dt=

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两边求导：-g(f(x))*f `(x)=1/2*{(x^1/2)}
注意到g(f(x))=x
f `(x)=-1/2*{x^(-1/2)}
======以下答案可供参考======
供参考答案1:
∫[1,f(x)]g(t)dt=-∫[0,1]g(t)dt+∫[0,f(t)g(t)dt
[∫[f(x),1]g(t)dt ] 'x=d∫[0,f(x)]g(t)dt/df(x) *df(x)/dx =(1/3)[x^(3/2)-8]'=(1/2)x^(1/2)
f(x)反函数是g(x)
d∫[0,f(x)]g(t)dt/df(x) =g(f(x))=x
x*df(x)/dx=(1/2)x^(1/2)
df(x)/dx=(1/2)x^(-1/2)