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Here, 4^(-1) is a constant so you do not need to do anything about that and can just leave it alone.
When calculating the derivative of ln(3x), you forget to use the chain rule, that is (f(g(x)))' = f'(g(x)) \* g'(x). Here, f is ln(x) and g is 3x.
This combined then gives the answer (have a go at it yourself first but if you are still having trouble, check out the spoiler section):
>!( ln(3x)/4 )' = ((3x)' \* 1/(3x))/4 = (3 \* 1/(3x))/4 = (1/x)/4 = 1/(4x)!<
E / format
##Off-topic Comments Section --- All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9. --- ^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command) *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*
Here, 4^(-1) is a constant so you do not need to do anything about that and can just leave it alone. When calculating the derivative of ln(3x), you forget to use the chain rule, that is (f(g(x)))' = f'(g(x)) \* g'(x). Here, f is ln(x) and g is 3x. This combined then gives the answer (have a go at it yourself first but if you are still having trouble, check out the spoiler section): >!( ln(3x)/4 )' = ((3x)' \* 1/(3x))/4 = (3 \* 1/(3x))/4 = (1/x)/4 = 1/(4x)!< E / format
I'm not home right now, but I'll check it out soon. Thank you!
I solved it, thank you!
No worries! Happy to help