derivative of log functions

[Guest]

How do you do this one, I cant seem to figure it out:

f(x) = ln [(x^2 + 1)/(x-1)]


and this one: h(u)= e^u squarerooted(lnusquarerooted)
I did this: e^u^1/2 ln u^1/2 but now I forget the rules, can u put the 1/2 infront of the e?

help, it would be lots of help if the steps were shown for these so I can recall the rules
Thanks for the help, and taking the time to answer.

 

[skeeter]

f(x) = ln [(x^2 + 1)/(x-1)]

first of all, "break" up the expression using the laws of logs ...

f(x) = log(x^2 + 1) - log(x - 1)

now, taking the derivative is as easy as falling off a ...

f'(x) = 2x/(x^2 + 1) - 1/(x - 1)

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for the second, do you mean this ... ?

\(\displaystyle \L h(u) = e^{\sqrt{u}*ln \sqrt{u}}\)

if so ... \(\displaystyle \L h(u) = e^{\sqrt{u}*ln \sqrt{u}} = \sqrt{u}*e^{\sqrt{u}}\)

\(\displaystyle \L h'(u) = \sqrt{u}*e^{\sqrt{u}}*\frac{1}{2\sqrt{u}} + e^{\sqrt{u}}*\frac{1}{2\sqrt{u}}\)

 

[Guest]

okay thanks, and for the 2nd one the ln u square rooted is on the bottom, its not part of the expodent, sorry for the confusion

 

[skeeter]

is this what you mean?

\(\displaystyle \L h(u) = \frac{e^{\sqrt{u}}}{ln\sqrt{u}}\)

 

[soroban]

Hello, bittersweet!

and this one: h(u)= e^u squarerooted(lnusquarerooted)
I did this: e^u^1/2 ln u^1/2 but now I forget the rules, can u put the 1/2 in front of the e?

I'm guessing that the function is: \(\displaystyle \L\,f(u)\;=\;e^{u^{\frac{1}{2}}}\,\cdot\,\ln\left(u^{\frac{1}{2}}\right)\)

This can simplified first: \(\displaystyle \:\,f(u) \;=\;e^{u^{\frac{1}{2}}}\,\cdot\frac{1}{2}\cdot\ln(u) \;= \;\frac{1}{2}\cdot e^{u^{\frac{1}{2}}}\cdot\ln(u)\)

Then: \(\displaystyle \L\,f'(u)\;=\;\frac{1}{2}\cdot e^{u^{\frac{1}{2}}}\cdot\frac{1}{u}\,+\,\frac{1}{2}\cdot\frac{1}{2}u^{-\frac{1}{2}}\cdot\ln(u) \;= \;\frac{e^{u^{\frac{1}{2}}}}{2u}\,+\,\frac{e^{u^{\frac{1}{2}}}\cdot\ln(u)}{4u^{\frac{1}{2}}}\)

Therefore: \(\displaystyle \L\,f'(u)\;=\;\frac{e^{u^{\frac{1}{2}}}}{4u}\left[2\,+\,u^{\frac{1}{2}}\cdot\ln(u)\right] \;=\;\frac{e^{\sqrt{u}}}{4u}\left[2\,+\,\sqrt{u}\cdot\ln(u)\right]\)