y= ln(3x+4)^4
i got
4 ln (3x+4) then = 4 * 1/ (3x+4) * 3
4 * 3/ (3x+4) y' = 12/(3x+4)
Ah, $\displaystyle \log(x^2) = 2\log(x) \neq (\log(x))^2$ which is where you've gone wrong.
Just differentiate it normally...
$\displaystyle y = (\log(3x + 4))^4.$
then $\displaystyle \frac{dy}{dx} = 4(\log(3x+4))^3 \cdot (\log(3x+4))'$
$\displaystyle = 4(\log(3x+4))^3 \cdot \frac{3}{3x+4} = \frac{12(\log(3x+4))^3}{3x+4}$
You missed a minus sign in the above, otherwise correct.
ln 2/x = -1/x
For your original question... Fine, but learn how to properly write it out as you'll get probably close to 0 for the above type of working. It makes no mathematical sense although I know what you are saying with it.
y = ln(3x+4)^4
= 4 * ln(3x+4)^3 * 3/(3x+4)
= 12 ln(3x+4)/(3x+4)
I'm sure you are aware that the derivative of x^n = nx^(n-1) * 1 since 1 is the derivative of x.
Same thing applies for functions.
Derivative of (some function)^n = n((some function)^(n-1) * derivative of (some function)
Or in maths notation.
$\displaystyle y = (f(x))^n$
$\displaystyle \frac{dy}{dx} = n(f(x))^{n-1} \cdot f'(x)$