1. log differntial

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)

2. Ah, $\log(x^2) = 2\log(x) \neq (\log(x))^2$ which is where you've gone wrong.

Just differentiate it normally...

$y = (\log(3x + 4))^4.$

then $\frac{dy}{dx} = 4(\log(3x+4))^3 \cdot (\log(3x+4))'$

$= 4(\log(3x+4))^3 \cdot \frac{3}{3x+4} = \frac{12(\log(3x+4))^3}{3x+4}$

3. another one

ln 2/x

= ln2 - ln x
ln 2= 1/2 *0 = 0
ln x = 1/x *1

so ln 2/x = 1/x ?
the question above can you do it in the form that i did much easy for me to understand.

4. Originally Posted by matlondon
another one

ln 2/x

= ln2 - ln x
ln 2= 1/2 *0 = 0
ln x = 1/x *1

so ln 2/x = 1/x ?
the question above can you do it in the form that i did much easy for me to understand.

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.

$y = (f(x))^n$

$\frac{dy}{dx} = n(f(x))^{n-1} \cdot f'(x)$

5. thx for the time