| ln(x2+1) | ||
logsinxx2+1 = | ||
| lnsinx |
| ln(x2+1) | (ln(x2+1))'*lnsinx−(lnsinx)'*ln(x2+1) | |||
( | )' = | (*) | ||
| lnsinx | ln2sinx |
| 1 | ||
Teraz liczmy (ln(x2+1))' podstawiamy x2+1=t i mamy (ln(x2+1))' = (ln(t))' = | * t' = | |
| t |
| 1 | 2x | |||
* 2x = | ||||
| x2+1 | x2+1 |
| 1 | 1 | |||
analogicznie (lnsinx)', t=sinx (lnt)' = | * t' = | * cosx = ctgx czyli: | ||
| t | sinx |
| 2xlnsinx/(x2+1) − ctgxln(x2+1) | ||
(*) = | ||
| ln2sinx |
