natural log derivative

[airken]

if f(x)= 3x lnx, then f'(x)=?

i used
f'(x)=3x(D lnx) + D (3x) (lnx)
f'(x)=3x (1/x) + 3 (lnx)
so... f'(x)=3+3lnx or 3(1+lnx).
unfortunately this isn't one of the possible answers given. could one of you kind folks help me understand where i went wrong?

thank you

 

[JeffM]

if f(x)= 3x lnx, then f'(x)=?

i used
f'(x)=3x(D lnx) + D (3x) (lnx)
f'(x)=3x (1/x) + 3 (lnx)
so... f'(x)=3+3lnx or 3(1+lnx).
unfortunately this isn't one of the possible answers given. could one of you kind folks help me understand where i went wrong?

thank you

What answers were you given to choose among?

Was one of them \(\displaystyle ln(e^3x^3)\ or\ ln(\{ex\}^3)\)?

 

[airken]

possible answers

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​Wow! that was sooo fast. thank you for the help.
sorry, i should have thought to provide the answers
here they are.

a) 3+ln(x^3)
b) 1+ln(x^3)
c) (3/x)+3lnx
d) 3/(x^2)
e) 1/x

 

[JeffM]

if f(x)= 3x lnx, then f'(x)=?

i used
f'(x)=3x(D lnx) + D (3x) (lnx)
f'(x)=3x (1/x) + 3 (lnx)
so... f'(x)=3+3lnx or 3(1+lnx).
unfortunately this isn't one of the possible answers given. could one of you kind folks help me understand where i went wrong?

thank you

You did not go wrong. You just did not go far enough for your book. This is one reason why I do not like multiple choice.

Answers to choose from:

a) 3+ln(x^3)
b) 1+ln(x^3)
c) (3/x)+3lnx
d) 3/(x^2)
e) 1/x

\(\displaystyle f'(x) = 3 + 3ln(x)\) is A correct answer but it implies by the laws of logarithms

\(\displaystyle f'(x) = 3 + ln(x^3)\), which is answer a in your book, but which further implies

\(\displaystyle f'(x) = 3 * 1 + ln(x^3) = 3ln(e) + ln(x^3) = ln(e^3) + ln(x^3) = ln(e^3x^3) = ln(\{ex\}^3).\)

There are lots of ways to express what is fundamentally the same answer.