# Differentiation of Log functions

• Jul 1st 2006, 07:34 PM
c_323_h
Differentiation of Log functions
Sigh...calculus is giving me a hard time...Trig was much easier. Anyone recommend a good calculus book to help me put things in perspective? Anyways, I need help with differentiating the following log functions:

1. $f(x)=\sqrt{x} lnx$
2. $f(u)=\frac{lnu}{1+ln(2u)}$

thanks
• Jul 1st 2006, 07:46 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
1. $f(x)=\sqrt{x} lnx$

First the product rule,
$(\sqrt{x})'\ln x+\sqrt{x}(\ln x)'$
Thus,
$\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}$
Thus,
$\frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt{x}}$
Thus,
$\frac{1}{\sqrt{x}} (1/2\ln x+1)$
• Jul 1st 2006, 07:50 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
Anyone recommend a good calculus book to help me put things in perspective?

The publishers that I love,
Prentice Hall and McGraw-Hill search their sites.
(The books are very expensive).

So you might be tempted to buy,
Calculus for Dummies or Schaum's Outline- I think these publishers are horrific and stay away from them.
• Jul 1st 2006, 10:33 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
The publishers that I love,
Prentice Hall and McGraw-Hill search their sites.
(The books are very expensive).

textboooks! you're freakin psycho! just kidding...i was thinking more along the lines of "A Tour of the Calculus". i used the product rule but my answer wasn't the same as the answer in my book..hmm maybe the book is wrong or i didn't simplify it correctly.
• Jul 1st 2006, 11:12 PM
Soroban
Hello, c_323_h!

I tried something on #2
. . and found it can be simplified beyond all recognition . . .

Quote:

$2)\;\;f(u)\;=\;\frac{\ln u}{1+\ln2u}$

Since $\ln(2u) \:=\:\ln(2) + \ln(u)$, we have:

. . $f(u)\;=\;\frac{\ln u }{1 + \ln2 + \ln u}$

Differentiate:

. . $f'(u) \;= \;\frac{[1 + \ln 2 + \ln u]\cdot\frac{1}{u} - \ln u\cdot\frac{1}{u}}{(1 + \ln2 + \ln u)^2}$

Multiply top and bottom by $u:$

. . $f'(u) \;= \;\frac{1 + \ln 2 + \ln u - \ln u}{u(1 + \ln 2 + \ln u)^2}$

Therefore: . $f'(u) \;=\;\frac{1 + \ln 2}{u(1 + \ln2u)^2}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let's see what they can do to #1 . . .

$1)\;\;f(x)\;=\;x^{\frac{1}{2}}\ln x$

Then: . $f'(x)\;=\;\left(\frac{1}{2}x^{-\frac{1}{2}}\right)\!\cdot\!\ln x \,+ \,x^{\frac{1}{2}}\!\cdot\!\left(\frac{1}{x}\right) \;=$ $\frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}
$

Get a common denominator: . $f'(x)\;=\;\frac{\ln x}{2\sqrt{x}} + \frac{2}{2\sqrt{x}} \;= \;\frac{\ln x + 2}{2\sqrt{x}}$

• Jul 1st 2006, 11:24 PM
c_323_h
Quote:

Originally Posted by Soroban
Hello, c_323_h!

I tried something on #2
. . and found it can be simplified beyond all recognition . . .

Since $\ln(2u) \:=\:\ln(2) + \ln(u)$, we have:

. . $f(u)\;=\;\frac{\ln u }{1 + \ln2 + \ln u}$

Differentiate:

. . $f'(u) \;= \;\frac{[1 + \ln 2 + \ln u]\cdot\frac{1}{u} - \ln u\cdot\frac{1}{u}}{(1 + \ln2 + \ln u)^2}$

Multiply top and bottom by $u:$

. . $f'(u) \;= \;\frac{1 + \ln 2 + \ln u - \ln u}{u(1 + \ln 2 + \ln u)^2}$

Therefore: . $f'(u) \;=\;\frac{1 + \ln 2}{u(1 + \ln2u)^2}$

THANK YOU! That's the exact answer as in my book. So that's how they got it! Does anyone know of any other ways to solve this one?
• Jul 2nd 2006, 12:40 AM
Soroban
Hello again, c_323_h!

Quote:

Does anyone know of any other ways to solve this one?

Since $1 + \ln(2u) \;=\;\ln(e) + \ln(2u) \;=\;\ln(2eu)$

. . we have: . $f(u)\:=\:\frac{\ln(u)}{\ln(2eu)}$

Quotient Rule: . $f'(u)\;=\;\frac{\ln(2eu)\cdot\frac{1}{u} - \ln(u)\cdot\frac{1}{2eu}\cdot2e}{[\ln(2eu)]^2}$

We have: . $f'(u)\;=\;\frac{\frac{\ln(2eu)}{u} - \frac{\ln(u)}{u}}{\ln(2eu)]^2}$

Multiply top and bottom by $u:\;\;f'(u) \;=\;\frac{\ln(2eu) - \ln(u)}{u[\ln(2eu)]^2}\;=$ $\frac{\ln\left(\frac{2eu}{u}\right)}{u[\ln(2eu)]^2}$

Therefore: . $f'(u) \;= \;\frac{\ln(2e)}{u[\ln(2eu)]^2]}$

• Jul 2nd 2006, 10:15 AM
c_323_h
thanks. what does differentiate with respect to x mean?
• Jul 2nd 2006, 10:48 AM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
thanks. what does differentiate with respect to x mean?

Nothing, just ignore that.
Basically, just differentiate the function.