Question

Find the derivative of ln(x^3+3x)^3

Answer 1

depends on what is being cubed

Answer 2

if it is
\[\ln((x^3+3x)^3)\] then the first step would be to rewrite it as
\[3\ln(x^3+3x)\]

Answer 3

f（x）=[3ln(x^3+3x)*(3x^2+x)]/(x^3+3x)

Answer 4

then it will be real easy, since
\[\frac{d}{dx}[\ln(f(x))]=\frac{f'(x)}{f(x)}\]

Answer 5

if the whole thing is being cubed, then the answer will be different

Answer 6

sorry, it should be f（x）=[3ln(x^3+3x)^2*(3x^2+x)]/(x^3+3x)

Answer 7

\[\ln(x^3+3x)^3\]

Answer 8

@satellite73 So, would it be (3(x^3+3x)^2+3x^2+3)/ln(x^3+3x)^3

Too many threes. @w@

Answer 9

it really depends on what is being raised to the power of 3, the input of the log, or the whole thing

Answer 10

that is, is it
\[\left(\ln(x^3+3x)\right)^3\] or is it
\[\ln((x^3+3x)^3)\]

Answer 11

if it is the first one, you get
\[3\ln(x^3+3x)^2\times \frac{3x^2+3}{x^3+3x}\]

Answer 12

It's the second thing-- the ln isn't cubed.

Answer 13

if it is the second one, you get \[3\times \frac{3x^2+3}{x^3+3x}\]

Answer 14

I tried to work it out and I got:\[\frac{ 9x^2+9 }{ (x^3+3x)^3 }\]

Answer 15

the denominator is wrong

Answer 16

Oh, it's just the inside function on the bottom?

Answer 17

yes

Answer 18

the three comes out front as a multiplier by the property of the log that says
\[\ln(x^n)=n\ln(x)\]