高等函数求导y=ln√((1+x)/(x-1))
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高等函数求导y=ln√((1+x)/(x-1))
![高等函数求导y=ln√((1+x)/(x-1))](/uploads/image/z/5139063-63-3.jpg?t=%E9%AB%98%E7%AD%89%E5%87%BD%E6%95%B0%E6%B1%82%E5%AF%BCy%3Dln%E2%88%9A%28%281%2Bx%29%2F%28x-1%29%29)
令t=√((1+x)/(x-1)),则y=ln t
y'=t'/t
t=√((1+x)/(x-1))=(1+ 2/(x-1) )^(1/2)
则t'=(1/2)·(1+ 2/(x-1) )^(-1/2) ·(-2/(x-1)²)
= -1/( t· (x-1)² )
因此
y'= -1/( t²· (x-1)² )
= -((x-1)/(1+x)) / (x-1)²
= 1/(1-x²)
y'=t'/t
t=√((1+x)/(x-1))=(1+ 2/(x-1) )^(1/2)
则t'=(1/2)·(1+ 2/(x-1) )^(-1/2) ·(-2/(x-1)²)
= -1/( t· (x-1)² )
因此
y'= -1/( t²· (x-1)² )
= -((x-1)/(1+x)) / (x-1)²
= 1/(1-x²)