# Quotient Rule Involving A Logarithm

• May 5th 2012, 02:42 PM
Bashyboy
Quotient Rule Involving A Logarithm
The problem is: $\displaystyle g(x) = \frac{10\log_{4} x}{x}$

By doing the quotient rule I get: $\displaystyle u = 10\log_{4} x$ and $\displaystyle u' = \frac{10}{ln 4\times{x}}$ and $\displaystyle v = x$ and $\displaystyle v' = 1$

When I apply the quotient rule I get: g'(x) = (10 - ln 4 log_4 x)/(x^2 ln 4) (sorry, I couldn't figure out the latex for this part.)

This answer is not even remotely close to the actual answer in the book. What did I do wrong?
• May 5th 2012, 02:56 PM
skeeter
Re: Quotient Rule Involving A Logarithm
Quote:

Originally Posted by Bashyboy
The problem is: $\displaystyle g(x) = \frac{10\log_{4} x}{x}$

By doing the quotient rule I get: $\displaystyle u = 10\log_{4} x$ and $\displaystyle u' = \frac{10}{ln 4\times{x}}$ and $\displaystyle v = x$ and $\displaystyle v' = 1$

When I apply the quotient rule I get: g'(x) = (10 - ln 4 log_4 x)/(x^2 ln 4) (sorry, I couldn't figure out the latex for this part.)

This answer is not even remotely close to the actual answer in the book. What did I do wrong?

note the change of base formula ...

$\displaystyle \log_b{a} = \frac{\ln{a}}{\ln{b}}$

$\displaystyle g(x) = \frac{10\log_{4} x}{x} = \frac{10}{\ln{4}} \cdot \frac{\ln{x}}{x}$

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