Cho hàm số \(f\left( x \right) = \frac{{{x^2}}}{{ - x + 1}}.\) Tìm \({f^{\left( {30} \right)}}\left( x \right).\)

* Hướng dẫn giải

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

\(f'\left( x \right) =  - 1 + \frac{{1!}}{{{{\left( {x - 1} \right)}^2}}};f''\left( x \right) = \frac{{2!}}{{{{\left( {x - 1} \right)}^3}}},{f^{\left( 3 \right)}} = \frac{{3!}}{{{{\left( {x - 1} \right)}^4}}} \Rightarrow {f^{\left( {30} \right)}} =  - \frac{{30!}}{{{{\left( {x - 1} \right)}^{31}}}} = \frac{{30!}}{{{{\left( {1 - x} \right)}^{31}}}}\)