设y=ln根号下1-x/1+x,则y''为多少
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设y=ln根号下1-x/1+x,则y''为多少
![设y=ln根号下1-x/1+x,则y''为多少](/uploads/image/z/8486971-43-1.jpg?t=%E8%AE%BEy%3Dln%E6%A0%B9%E5%8F%B7%E4%B8%8B1-x%2F1%2Bx%2C%E5%88%99y%27%27%E4%B8%BA%E5%A4%9A%E5%B0%91)
y=ln根号下1-x/1+x
=0.5ln(1-x)-0.5ln(1+x)
y'=0.5/(1-x) - 0.5/(1+x)
=0.5(1+x-1+x)/(1-x)(1+x)
=x/(1-x²)
y''=(1-x²)+2x²/(1-x²)²
=(1+x²)/(1-x²)²
再问: y=ln根号下1-x/1+x =0.5ln(1-x)-0.5ln(1+x) y'=0.5/(1-x) - 0.5/(1+x)这里不应该是复合函数求导=-0.5/(1-x) - 0.5/(1+x)? =0.5(1+x-1+x)/(1-x)(1+x) =x/(1-x²)
再答: y'=-0.5/(1-x) - 0.5/(1+x) =0.5(-1-x-1+x)/(1-x)(1+x) =-1/(1-x²) =1/(x²-1) y'‘=-2x/(x²-1)²
=0.5ln(1-x)-0.5ln(1+x)
y'=0.5/(1-x) - 0.5/(1+x)
=0.5(1+x-1+x)/(1-x)(1+x)
=x/(1-x²)
y''=(1-x²)+2x²/(1-x²)²
=(1+x²)/(1-x²)²
再问: y=ln根号下1-x/1+x =0.5ln(1-x)-0.5ln(1+x) y'=0.5/(1-x) - 0.5/(1+x)这里不应该是复合函数求导=-0.5/(1-x) - 0.5/(1+x)? =0.5(1+x-1+x)/(1-x)(1+x) =x/(1-x²)
再答: y'=-0.5/(1-x) - 0.5/(1+x) =0.5(-1-x-1+x)/(1-x)(1+x) =-1/(1-x²) =1/(x²-1) y'‘=-2x/(x²-1)²