| x | |
| x2+1 |
| f | f'*g−f*g' | |||
( | )'= | |||
| g | g2 |
| −x2+1 | ||
w 1 pochodnej wyszło mi | czy to jest dobrze? | |
| (x2+1)2 |
.
| −2x(x2+1)2+x2−1(x2+1)*2*2x | ||
| (x2+1)4 |
| 2x5+4x3−6x | ||
a po wyliczeniu mam | ||
| (x2+1)4 |
| −2x(x2+1)2 − 2(x2+1)*2x*(−x2+1) | ||
f"(x) = | = | |
| (x2+1)4 |
| (x2+1)*[ −2x(x2+1) − 2*2x*(−x2+1) ] | |
= | |
| (x2+1)4 |
| −2x3 − 2x + 4x2 − 4x | |
= | |
| (x2+1)3 |
| −2x(x2 − 2x + 3) | |
| (x2+1)3 |