Help with differentiation of natural logarithm Find $\;\dfrac{dy}{dx}\;$ given $y=\frac{\ln(8x)}{8x}$.  
The answer is $\;\dfrac{1-\ln(8x)}{8x^2}\;$.  
Can you show the process of how this is worked?  
Thanks.
 A: Here we can use the quotient rule and the chain rule:
Quotient rule: $\quad$ If $y = \dfrac{f(x)}{g(x)}$, then $$\dfrac{dy}{dx} = \dfrac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\tag{(1) quotient rule}$$
$\displaystyle f(x) = \ln(8x), \; f'(x) = \frac 1{8x}\cdot \frac d{dx}(8x) =  \dfrac 8{8x} = \dfrac 1x\quad\text{by chain rule}$
$\displaystyle  g(x) = 8x, \; g'(x) = 8.$
Now, just substitute each of $f(x), g(x), f'(x), g'(x), \text{and} g^2(x) = (8x)^2$ into $(1)$, simplify, and you're done!
$$\frac {dy}{dx} = \frac{(8x)\frac{1}{x} - \left(\ln(8x)\right)\cdot 8}{64x^2} = \dfrac{(1 - \ln(8x))}{8x^2}$$
A: There are two approaches here.  The most straightforward way is to apply the quotient rule.
That is, given $f(x) = \frac{g(x)}{h(x)}$:
$$f'(x) = \frac{g'(x)\cdot h(x) - g(x)\cdot h'(x)}{(h(x))^2}$$
Or, in words: "bottom times the derivative of the top, minus the top times the derivative of the bottom; all over the bottom squared."
So, we have that the "top" is $\ln 8x$.  The derivative of that is $\frac{1}{x}$, applying the chain rule and the derivative of natural log.
The bottom is $8x$.  The derivative of this is $8$.  (just a power rule)
We plug into the quotient rule:
$$f'(x) = \frac{(8x)\frac{1}{x} - \Big(\ln(8x)\Big)8}{(8x)^2}$$
Simplifying:
$$f'(x) = \frac{8 - 8\ln(8x)}{(8x)^2}$$
$$f'(x) = \frac{1 - \ln(8x)}{8x^2}$$
This is the correct answer.
EDIT: I forgot I mentioned that there were two approaches!  The other approach is to re-write the problem using properties of exponents, and apply the product rule:
$$f(x) = \frac{1}{8}x^{-1}\ln(8x)$$
(Differentiate with the product rule--it's a good practice.  You should end up with the same function as above.)
A: $\frac{d(ln(x))}{dx}=\frac{1}{x}$, you mean this?
A: This is just the quotient rule: $$\frac{1}{8}\left[\frac{x\cdot \frac{1}{x}-1\cdot\ln(8x)}{x^2}\right].$$
A: $$y=\frac{\ln(8x)}{8x}$$
$$\frac{dy}{dx} = \frac{8x \frac{d(\ln(8x))}{dx} - \ln(8x) \frac{d(8x)}{dx}}{(8x)^2}$$
Then use the following facts:
$$\frac{d \ln(ax)}{dx}=\frac{a}{ax}$$
and 
$$\frac{d (ax)}{dx}=a$$
A: $h(x)=\frac{\ln(8x)}{8x}$.
Use the Quotient Rule. That is given $\frac{f(x)}{g(x)}$, its derivative is $\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}$
$h'(x)=\frac{(1/x)(8x)-(8)(\ln(8x))}{64x^2}$
$=\frac{8-8\ln8x}{64x^2}=\frac{8(1-\ln8x)}{64x^2}=\frac{1-\ln x}{8x^2}$
