Question

d/dx ln[x(x^2+3)^3] = ?
How to solve this?

Answer 1

\[\frac{ d }{ dx } \ln \left[x \left( x^2+3 \right)^3 \right]\] is this the function?

Answer 2

yes

Answer 3

so\[\frac{ d }{ dx}\ln u = \frac{u'}{ u }\]

Answer 4

you can multiply or use product rule for x(x^2+3)^3

Answer 5

I already tried but couldn't get the right result

Answer 6

\[\frac{(x ^{2}+3)^{3} +3(x^{2}+3)^{2}2x}{x(x^{2}+3)^{3}}\]

Answer 7

is this right?

Answer 8

no.
\[u = x(x^2+3)^3 \Rightarrow u' = (x^2+3)^3+x(2)(x^2+3)^2(2x)=(x^2+3)^2(x^2+3+4x^2)\]
\[=(5x^2+3)(x^2+3)^2\]

Answer 9

Hmm....
\[(x^{2}+3)^{3} + x(3)(x^{2}+3)^{2}(2x) = (x^{2}+3)^{2}[(x^{2}+3)^{1}+ 6x^{2}] = (x^{2}+3)^{2}(7x^{2}+3)\]
\[\frac{ (x^{2}+3)^{2}(7x^{2}+3) }{ x(x^{2}+3)^{3} }\implies \frac{ (7x^{2}+3) }{ x(x^{2}+3) }\]

Answer 10

Aww, went off the edge

Answer 11

oops, should be 3x^2 giving just what @psymon wrote.

Answer 12

ah thank you, I think the answer of this question is wrong, it is (1-x^2)/x(x^2+1), which is not related at all =.=

Answer 13

lol.

Answer 14

where did that come from?

Answer 15

my homework from my professor

Answer 16

the solution? perhaps your prof was looking at a different prob...

Answer 17

yep, i think so, he sometimes makes mistakes at class xD

Answer 18

it happens... it's so difficult to be perfect! in fact, i'd say impossible.