# Differentiation of Log functions

• Jul 1st 2006, 06:34 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. $\displaystyle f(x)=\sqrt{x} lnx$
2. $\displaystyle f(u)=\frac{lnu}{1+ln(2u)}$

thanks
• Jul 1st 2006, 06:46 PM
ThePerfectHacker
Quote:

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

First the product rule,
$\displaystyle (\sqrt{x})'\ln x+\sqrt{x}(\ln x)'$
Thus,
$\displaystyle \frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}$
Thus,
$\displaystyle \frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt{x}}$
Thus,
$\displaystyle \frac{1}{\sqrt{x}} (1/2\ln x+1)$
• Jul 1st 2006, 06:50 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, 09:33 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, 10:12 PM
Soroban
Hello, c_323_h!

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

Quote:

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

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

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

Differentiate:

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

Multiply top and bottom by $\displaystyle u:$

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

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

If that's what they did, no wonder your answer is different!

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

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

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

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

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

• Jul 1st 2006, 10:24 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 $\displaystyle \ln(2u) \:=\:\ln(2) + \ln(u)$, we have:

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

Differentiate:

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

Multiply top and bottom by $\displaystyle u:$

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

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

If that's what they did, no wonder your answer is different!

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 1st 2006, 11:40 PM
Soroban
Hello again, c_323_h!

Quote:

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

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

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

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

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

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

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

• Jul 2nd 2006, 09:15 AM
c_323_h
thanks. what does differentiate with respect to x mean?
• Jul 2nd 2006, 09:48 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.