Question

Cho hàm số \(f(x) = \frac{{x + 1}}{{x - 1}}\). Tính \(f''(1)\).

Answer 1

\(f'\left(x\right)=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}\)

\(=\dfrac{x-1-x-1}{\left(x-1\right)^2}=\dfrac{-2}{\left(x-1\right)^2}\)

=>\(f''\left(x\right)=\dfrac{-2\cdot\left(-2\right)\left(x-1\right)}{\left(x-1\right)^4}=\dfrac{4}{\left(x-1\right)^3}\)