- Thread starter akleron
- Start date

\(\displaystyle \dfrac{d}{ds}e^s = e^s\)

\(\displaystyle \dfrac{d}{ds} s \ln(s) = \ln(s) + 1\)

\(\displaystyle \dfrac{d}{ds}\dfrac{e^s}{s\ln(s)}= \dfrac{e^s(s \ln(s)) - e^s(\ln(s)+1)}{(s\ln(s))^2} = \\~\\

\dfrac{e^s}{s \ln(s)} - \dfrac{e^s}{s^2\ln(s)} - \dfrac{e^s}{s^2(\ln(s))^2}

\)

you seemed to miss the fact that \(\displaystyle \dfrac{d}{ds} s = 1\)

Lol should've know that (s)'=1 ...

\(\displaystyle \dfrac{d}{ds}e^s = e^s\)

\(\displaystyle \dfrac{d}{ds} s \ln(s) = \ln(s) + 1\)

\(\displaystyle \dfrac{d}{ds}\dfrac{e^s}{s\ln(s)}= \dfrac{e^s(s \ln(s)) - e^s(\ln(s)+1)}{(s\ln(s))^2} = \\~\\

\dfrac{e^s}{s \ln(s)} - \dfrac{e^s}{s^2\ln(s)} - \dfrac{e^s}{s^2(\ln(s))^2}

\)

you seemed to miss the fact that \(\displaystyle \dfrac{d}{ds} s = 1\)

Thank you !

- Joined
- Mar 16, 2016

- Messages
- 1,958

Also, should be -1 not +1 in square bracket.

- Joined
- Jan 27, 2012

- Messages
- 6,293