super
lubie to: Obliczyc pochodna na zaraz :
y = tg2√1/x
9 mar 09:16
danoel:
| | 1 | |
= 2 * |
| * ( √x−1 )' = |
| | 1+x−1 | |
| | 1 | | 1 | |
2 * |
| * |
| * (x−1)' = |
| | 1+x−1 | | 2√x do potegi(−1) | |
| | 1 | | 1 | |
2 * |
| * |
| * (−1) * x−2 |
| | 1+x−1 | | 2√x do potegi(−1) | |
9 mar 09:49
danoel: sory oczywiście to jst zle pomylilem pochodna tg z ctg
9 mar 09:57
danoel: sory pomylilem tg z arc tg
9 mar 09:58
danoel: może lepiej jak policzy to ktoś mądrzejszy
9 mar 10:00
huehuehue: | | 1 | | −1 | |
niech t=√ |
| t'= |
| |
| | x | | 2x2√1/x | |
| | 2tgt | |
y'=( |
| )*t' podstaw sobie teraz t i t' |
| | cos2t | |
9 mar 12:07