# Thread: Derivatives of Logarithmic and Exponential Function

1. ## Derivatives of Logarithmic and Exponential Function

y = (1+ log(x-numerator)3-base) / x

I didnt know how to write the log but the numerator is x and the base is 3.

2. Originally Posted by rhhs11
y = (1+ log(x-numerator)3-base) / x

I didnt know how to write the log but the numerator is x and the base is 3.

Before I try anything, let me make sure I have read your explanation right...

$\displaystyle f(x)=\frac{1+\log_3{x}}{x}$

Is this what you meant?

3. Everything is over including 1. So its (1+log..) / x

Thanks

4. $\displaystyle f(x)=\frac{1+\log_3{x}}{x}$

First lets use the change of base property to get rid of the $\displaystyle \log_3{x}$

$\displaystyle \log_3{x}=\frac{\ln{x}}{\ln{3}}$

Substituting this we get,

$\displaystyle f(x)=\frac{1+\frac{\ln{x}}{\ln{3}}}{x}$

Now to diferentiate this you need to either bring up the x and use the product rule or use the quotient rule.

$\displaystyle \frac{d}{dx}f(x)=\left(1+ \frac{1}{\ln{3}}\ln{x}\right)(-x^{-2})+(x^{-1})\left( \frac{1}{\ln{x}}\left(\frac{1}{x}\right)\right)$

Doing several steps of simplification and using a sum of logarithms property we get...

$\displaystyle \frac{d}{dx}f(x)=\frac{\ln{3x}-1}{x^2\ln{3}}$