sorry to ask a second question so soon, but i'm just not getting this one.

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 that isn't one of the possible answers given. could one of you kind folks help me understand where i went wrong?

thank you

I am getting the same answer.

What are the other choices?

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

It is a. They used the law of logs or lns to rewrite 3 ln x as ln x^3.

Thank you very much. haven't touched this kind of stuff in a while. your help is much appreciated.

You are welcome. So many times, students know the Calculus, but it is something from past math courses that cause the issue with finding the final answer.

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No problem at all! Let's go through this step-by-step to identify where you went wrong.

To find the derivative of f(x) = 3x lnx, you applied the product rule correctly, but made a mistake when taking the derivative of lnx.

Let's start from the beginning and apply the product rule correctly:

f(x) = 3x lnx

To differentiate this expression, we use the product rule:

f'(x) = (3x) * (D lnx) + (lnx) * (D 3x)

Now, let's take the derivative of lnx:

Using the chain rule, we have D lnx = 1/x

Now, substitute back into the product rule formula:

f'(x) = (3x) * (1/x) + (lnx) * (D 3x)

Simplifying further:

f'(x) = 3 + 3lnx

So, the correct derivative of f(x) is f'(x) = 3 + 3lnx.

This is one of the possible answers given, so you did not go wrong there. It seems like you just made a mistake when differentiating lnx.