Question

y=ln((x^+3X)/(X-1))

Answer 1

derivative

Answer 2

are you familiar with the derivative of ln u ? if so, you can apply the chain rule

Answer 3

also, could you clarify what is inside the first parentheses ? x^ +3x ?

Answer 4

(4x^3/(x^4+3x))-(1/x-1) ?

Answer 5

No is it ((4x^3+3)/(x^4+3x))-(1/x-1) ?

Answer 6

i am not sure what is inside that first parentheses ?

Answer 7

you wrote

y=ln((x^+3X)/(X-1))

what is the exponent on the first x?

Answer 8

ups: y=ln((x^4+3X)/(X-1))

Answer 9

you may want to divide this into two separate problems

remember

\[\log(\frac{ a }{ b }) = \log (a) - \log (b)\]
our problem simplifies into
\[y = \log(x^4 + 3x) - \log(x - 1)\]
did you get this far ?

Answer 10

thank you for your time ( this helped a lot :) I will look at it to nite

Answer 11

good luck