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)

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- Apr 10th 2010, 11:59 AMmatlondonlog 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) - Apr 10th 2010, 12:09 PMDeadstar
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 $\displaystyle \frac{dy}{dx} = 4(\log(3x+4))^3 \cdot (\log(3x+4))'$

$\displaystyle = 4(\log(3x+4))^3 \cdot \frac{3}{3x+4} = \frac{12(\log(3x+4))^3}{3x+4}$ - Apr 10th 2010, 12:17 PMmatlondon
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. - Apr 10th 2010, 12:27 PMDeadstar
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.

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

$\displaystyle \frac{dy}{dx} = n(f(x))^{n-1} \cdot f'(x)$ - Apr 10th 2010, 01:01 PMmatlondon
thx for the time(Nod)