# Thread: Differentiation of Log functions

1. ## 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

2. 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,
$\displaystyle \frac{1}{\sqrt{x}} (1/2\ln x+1)$

3. 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.

4. 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.

5. Hello, c_323_h!

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

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

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

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

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

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

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?

7. Hello again, c_323_h!

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

8. thanks. what does differentiate with respect to x mean?

9. Originally Posted by c_323_h
thanks. what does differentiate with respect to x mean?
Nothing, just ignore that.
Basically, just differentiate the function.