Bài 9.28 trang 98

Đề bài

Cho hàm số $f\left(x\right)=\frac{x+1}{x-1}$. Tính $f^{\prime \prime }\left(0\right)$.

Lời giải chi tiết

$f^{\prime }\left(x\right)={\left(\frac{x+1}{x-1}\right)}^{\prime }=\frac{{\left(x+1\right)}^{\prime }\left(x-1\right)-\left(x+1\right){\left(x-1\right)}^{\prime }}{{\left(x-1\right)}^2}$

$=\frac{1.\left(x-1\right)-\left(x+1\right).1}{{\left(x-1\right)}^2}=\frac{x-1-x-1}{{\left(x-1\right)}^2}=\frac{-2}{{\left(x-1\right)}^2}$.

$f^{\prime \prime }\left(x\right)={\left[\frac{-2}{{\left(x-1\right)}^2}\right]}^{\prime }=\frac{{\left(-2\right)}^{\prime }{\left(x-1\right)}^2-\left(-2\right){\left[{\left(x-1\right)}^2\right]}^{\prime }}{{\left[{\left(x-1\right)}^2\right]}^2}$

$=\frac{0-\left(-2\right).2\left(x-1\right)}{{\left(x-1\right)}^4}=\frac{4\left(x-1\right)}{{\left(x-1\right)}^4}=\frac{4}{{\left(x-1\right)}^3}$.

$f^{\prime \prime }\left(0\right)=\frac{4}{{\left(0-1\right)}^3}=-4$.