f(x)=1/x
根据定义(f(x+x')-f(x))/x',x'->0
即f'(x)=（1/（x+x')-1/x)/x',x'->0
通分得到f'(x)=-1/（x^2+x*x'),x'->0
所以f'(x)=-1/x^2
(y1)'=
(y2)'=6x+2
(y3)'=
(y4)'=3*t^2 - 3^t*Ln(3)
(y5)'=(3x)/(a^2 - x^2)^(3/2)
