# Thread: Bases other than e... help?

1. ## 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

2. 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

3. ahh sorry, it says to differentiate the function.

4. 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

5. 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?

6. 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

7. oh wait, so would i get the derivative of lnx/ln10 first then plug it into that?

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

9. 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 ?

10. 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!)

11. 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

12. Hello, runner07!

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

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: .$\displaystyle \frac{1}{\ln10}\!\cdot\!\frac{1}{x}$

13. 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

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