Đề bài: 1) Cho $y = \frac{{{x^2}}}{{1 – x}}$, tìm $y^{(8)}$2) Cho $y = \frac{1}{{{x^2} – 3x + 2}}$, tìm ${y^{(n)}}$.
Lời giải
1) Ta có:
$y = \frac{{{x^2} – 1 + 1}}{{1 – x}} = – x – 1 – \frac{1}{{x – 1}} = – x – 1 – {(x – 1)^{ – 1}}$
$ \Rightarrow y’ = 1 – ( – 1).{(x – 1)^2}$
$y” = – ( – 1).( – 2){(x – 1)^3} = {( – 1)^3}.2!{(x – 1)^{ – 3}}$
$ \Rightarrow y”’ = {( – 1)^4}.3{(x – 1)^{ – 4}},…,\\
{y^{(8)}} = {( – 1)^9}.8!{(x – 1)^{ – 9}} = – \frac{{8!}}{{{{(x – 1)}^9}}}$
2) Ta có:
$y = \frac{1}{{(x – 1)(x – 2)}} = \frac{1}{{x – 2}} – \frac{1}{{x – 1}} = {(x – 2)^{ – 1}} – {(x – 1)^{ – 1}}$
Tương tự bài trên dẫn tới kết quả sau: ${y^{(n)}} = {( – 1)^n}.n!.\left[ {\frac{1}{{{{(x – 2)}^{n + 1}}}} – \frac{1}{{{{(x – 1)}^{n + 1}}}}} \right]$
Trả lời