find the derivative of f(x)=ln(x^{3}+3x)^{3}

[wind]

Hi I need help with this question, for my exam tomarrow...

find the derivative of \(\displaystyle \L\ f(x)=ln(x^{3}+3x)^{3}\)


\(\displaystyle \L\ f'(x)=\frac{1}{(x^{3}+3x)^{3}}*[ 3(x^{3}+3x)*3x+3]\)


\(\displaystyle \L\ f'(x)=\frac{(3x^{3}+9x)(3x+3)}{(x^{3}+3x)^{3}}\)


\(\displaystyle \L\ f'(x)=\frac{3(x^{3}+3x)(3x+3)}{(x^{3}+3x)^{3}}\)


\(\displaystyle \L\ f'(x)=\frac{9(x+1)}{(x^{3}+3x)^{2}}\)

what did I do wrong? Thanks

 

[skeeter]

I have no idea what you are doing.

the form of function that you have here is ...

\(\displaystyle \L \frac{d}{dx} \left[\ln{u^k}\right] = \frac{d}{dx} \left[k \cdot \ln{u}\right] = k \cdot \frac{u'}{u}\)

so ...

\(\displaystyle \L f(x) = \ln{(x^3 + 3x)^3}\)

\(\displaystyle \L f(x) = 3 \ln{(x^3 + 3x)}\)

\(\displaystyle \L f'(x) = 3 \cdot \frac{3x^2 + 3}{x^3 + 3x}\)

\(\displaystyle \L f'(x) = \frac{9(x^2 + 1)}{x(x^2 + 3)}\)

 

[galactus]

Chain rule:

\(\displaystyle \L\\\frac{d}{dx}[ln(x^{3}+3x)^{3}]dx\)

\(\displaystyle \L\\\underbrace{3ln(x^{3}+3x)^{2}}_{\text{outside}}\cdot\underbrace{\frac{1}{x^{3}+3x}}_{\text{derivative of\\lnx}}\cdot\overbrace{{3x^{2}+3}}^{\text{inside}}\)

\(\displaystyle \L\\f'(x)=ln(x^{3}+3x)^{2}\cdot\frac{9(x^{2}+1)}{x(x^{2}+3)}\)

 

[skeeter]

wait a moment ... is the original function

\(\displaystyle \L f(x) = \ln{(x^3 + 3x)^3}\) (as was posted originally)

or did you mean

\(\displaystyle \L f(x) = [\ln{(x^3+3x)}]^3\)

???

 

[wind]

Thanks skeeter, galactus

I have no idea what you are doing.

is'nt the derivitive of ln(fx)=1/f(x)*f'(x)

thats what i was trying to do, but I forgot to write \(\displaystyle \L\ 3x^{2}\) and probable made some other mistakes....

skeeter, it was the first one, posted originally, so ur solution is right.

 

[galactus]

\(\displaystyle ln(x^{3}+3x)^{3}\) is the same as \(\displaystyle (ln(x^{3}+3x))^{3}\)

I ran them through Maple and my TI for a check and I keep getting the answer I posted.

 

[skeeter]

don't know about Maple, but the TI interprets ln(u)² as [ln(u)]².

in normal, people-written notation, ln(u)² implies that only the "u" is being squared, hence
ln(u)² = 2ln(u).

try taking the derivative using this function syntax ... ln((x³+3x)³)

 

[tkhunny]

Lesson: Notation will save you. Never write an ambiguous expression. If we have to "interpret", we MAY get what you intend. Use whatever it takes to write clearly, so that no one can misunderstand.