# Bases other than e... help?

• Aug 30th 2007, 08:04 PM
runner07
Bases other than e... help?
im stuck here, how do i do this?

g(x) = ( log [10] x ) / (x^2)

note:
log [10] x-- this is log, sub 10, x... i didnt know how to write it out
• Aug 30th 2007, 08:05 PM
Jhevon
Quote:

Originally Posted by runner07
im stuck here, how do i do this?

g(x) = ( log [10] x ) / (x^2)

note:
log [10] x-- this is log, sub 10, x... i didn't know how to write it out

um, you didn't tell us what to do
• Aug 30th 2007, 08:13 PM
runner07
ahh sorry, it says to differentiate the function.
• Aug 30th 2007, 08:20 PM
Jhevon
Quote:

Originally Posted by runner07
ahh sorry, it says to differentiate the function.

ok. use the change of base formula for logarithms

recall that: $\displaystyle \log_a b = \frac {\log_c b}{\log_c a}$

so we can write: $\displaystyle \log_{10} x = \frac {\ln x}{\ln 10}$

now continue
• Aug 30th 2007, 08:25 PM
runner07
Quote:

Originally Posted by Jhevon
ok. use the change of base formula for logarithms

recall that: $\displaystyle \log_a b = \frac {\log_c b}{\log_c a}$

so we can write: $\displaystyle \log_{10} x = \frac {\ln x}{\ln 10}$

now continue

so then i would use the quotient rule? making it...

(x^2)(lnx/ln10) - (lnx/ln10)(2x)
(x^2)^2

is that right?
• Aug 30th 2007, 08:27 PM
Jhevon
Quote:

Originally Posted by runner07
so then i would use the quotient rule? making it...

(x^2)(lnx/ln10) - (lnx/ln10)(2x)
(x^2)^2

is that right?

no, that's not the quotient rule. try again
• Aug 30th 2007, 08:40 PM
runner07
oh wait, so would i get the derivative of lnx/ln10 first then plug it into that?
• Aug 30th 2007, 08:43 PM
Jhevon
Quote:

Originally Posted by runner07
oh wait, so would i get the derivative of lnx/ln10 first then plug it into that?

yes, remember the quotient rule: $\displaystyle \left( \frac {u}{v} \right)^{\prime} = \frac {u'v - v'u}{v^2}$

also, it may be easier for you to write: $\displaystyle g(x) = \frac {1}{\ln 10} \cdot \frac {\ln x}{x^2}$
• Aug 30th 2007, 08:48 PM
runner07
Quote:

Originally Posted by Jhevon
yes, remember the quotient rule: $\displaystyle \left( \frac {u}{v} \right)^{\prime} = \frac {u'v - v'u}{v^2}$

also, it may be easier for you to write: $\displaystyle g(x) = \frac {1}{\ln 10} \cdot \frac {\ln x}{x^2}$

so first i would do the quotient rule for the lnx/x^2 right? and then the product rule with the 1/ln10 ?
• Aug 30th 2007, 09:21 PM
earboth
Quote:

Originally Posted by runner07
so first i would do the quotient rule for the lnx/x^2 right? and then the product rule with the 1/ln10 ?

Hello,

not quite. Remember $\displaystyle \frac{1}{\ln(10)}$ is a constant.

$\displaystyle g(x) = \frac {1}{\ln 10} \cdot \frac {\ln x}{x^2}$ will give:

$\displaystyle g'(x) = \frac {1}{\ln 10} \cdot \frac {x^2 \cdot \frac{1}{x} - \ln (x) \cdot 2x}{x^4}$

I'll leave the rest for you. (You can cancel out a factor - under certain conditions!)
• Aug 31st 2007, 03:52 AM
Jhevon
Quote:

Originally Posted by runner07
so first i would do the quotient rule for the lnx/x^2 right? and then the product rule with the 1/ln10 ?

as earboth already said, when we have a constant multiplier to a derivative, we can essentially forget about it and differentiate the (variable) function and then apply the constant multiplier when we're done. in fact, this is one of the properties of derivatives:

$\displaystyle \frac {d}{dx} c \cdot f(x) = c \frac {d}{dx}f(x)$

where $\displaystyle c$ is a constant
• Aug 31st 2007, 06:19 AM
Soroban
Hello, runner07!

You should have been shown the formula for "other bases".

Quote:

im stuck here, how do i differentiate this?
. . $\displaystyle g(x) \:= \:\frac{\log_{10}x}{x^2}$

You can always derive the formula . . . (Yeah, right!)

We have: .$\displaystyle y \:=\:\log_b(x)$ . ... where $\displaystyle b$ is a positive real number $\displaystyle \neq 1$

. . Re-write in exponential form: .$\displaystyle b^y \:=\:x$

. . Take logs: .$\displaystyle \ln\left(b^y\right) \:=\:\ln(x)\quad\Rightarrow\quad y\!\cdot\!\ln(b) \:=\:\ln(x)$

. . Differentiate implicitly: .$\displaystyle \frac{dy}{dx}\!\cdot\ln(b) \:=\:\frac{1}{x}$

Therefore: . $\displaystyle \boxed{\frac{dy}{dx} \;=\;\frac{1}{\ln b}\!\cdot\!\frac{1}{x}}$

So the derivative of $\displaystyle \log_{10}x$ is: .$\displaystyle \frac{1}{\ln10}\!\cdot\!\frac{1}{x}$

• Aug 31st 2007, 06:48 AM
Krizalid
You can use the product rule too :)

$\displaystyle g(x)=\frac{\log_{10}x}{x^2}=\frac{\dfrac{\ln x}{\ln10}}{x^2}=\frac1{\ln10}\cdot\frac{\ln x}{x^2}$

So, all we have to do it's take the derivative of $\displaystyle \frac{\ln x}{x^2}=x^{-2}\ln x$, now by the product rule

$\displaystyle \ln 10\cdot g'(x)=-2x^{-3}\ln x+x^{-2}\cdot\frac1x=-\frac{2\ln x}{x^3}+\frac1{x^3}=\frac{1-2\ln x}{x^3}$