pochodna
manhattan: oblicz pochodną
y=(1+ 1/x )ln1/x
30 lip 21:49
pigor: ... logarytmując i różniczkując obustronnie np.
tak :
y= (1+
1x)
ln 1x= (1+
1x)
ln1−ln x ⇒ lny = ln(1+
1x)
−ln x ⇒
| | 1 | | 1 | | 1 | |
lny=−lnx*ln(1+1x) ⇒ |
| y'= −1x*ln(1+1x)−lnx* |
| *(− |
| ) ⇒ |
| | y | | 1+1x | | x2 | |
| | 1 | | lnx | |
⇒ |
| y' = −1x*ln(1+1x) + |
| /* y ⇒ |
| | y | | x(x+1) | |
⇒ y' =
1xy [
1x+1lnx − ln(1+
1x] ⇒
⇒
y' = 1x(1+
1x)
ln 1x [
1x+1lnx − ln(1+
1x] . ...
30 lip 22:28
konrad:
=eln (1+1/x)ln 1/x=
eln 1/x * ln (1+1/x)=
eln 1/x * ln (1+1/x)*(1/(1/x)*(−1/x2)*ln(1+1/x)+ln(1/x)*(1/(1+1/x))*(−1/x2)))=
(1+ 1/x )ln 1/x*(−1/x*ln(1+1/x)−ln(1/x)*1/(x2+x))=
−(1+ 1/x )ln 1/x*(ln(1+1/x)/x+ln(1/x)/(x(x+1)))=
−(1+ 1/x )ln 1/x*((x+1)ln(1+1/x)/(x(x+1))+ln(1/x)/(x(x+1)))=
−(1+ 1/x )ln 1/x*((x+1)ln(1+1/x)+ln(1/x))/(x(x+1))=
30 lip 22:45