Question

Can anyone get the derivative of this ?

y = 1/3 Log e^(3x+4) + Log e^(6x+8)

Answer 1

Are those logs of base 10?
Or is it like this?
\(\Large\rm \log_e (3x+4)\)
The notation is kind of weird... so I can't tell..

Answer 2

It a relief you'd answer me .

Answer 3

Need clarification on the format.

This: \(\Large\rm y=\frac{1}{3}\log e^{3x+4}+\log e^{6x+8}\)

Or this: \(\Large\rm y=\frac{1}{3}\log_e(3x+4)+\log_e(6x+8)\)

Which is the correct format?
The "^" is confusing.

Answer 4

If its ok Can you solve them Both . My teacher did not clarify it thats why when I simplified it to In(3x+4) +In(6x+8) they said it is wrong

Answer 5

Let's start with the second one, that one makes more sense.\[\Large\rm y=\frac{1}{3}\log_e(3x+4)+\log_e(6x+8)\]\[\Large\rm y=\frac{1}{3}\ln(3x+4)+\ln(6x+8)\]Recall the derivative of ln(x) is 1/x.
Same idea here, derivative of ln(stuff) is 1/(stuff) then we have to remember to apply the chain rule as well.

Answer 6

Yeah That What I did . Then I got The answer y ' =1/ (3x+4) + 6/ (6x+8)

Answer 7

\[\Large\rm y'=\frac{1}{3}\frac{1}{3x+4}\color{royalblue}{(3x+4)'}+\frac{1}{6x+8}\color{royalblue}{(6x+8)'}\]
\[\Large\rm y'=\frac{1}{3}\frac{1}{3x+4}\color{royalblue}{(3)}+\frac{1}{6x+8}\color{royalblue}{(6)}\]And that was incorrect?
Ok let's try the other form then.

Answer 8

We'll assume these are logs of base 10,\[\Large\rm y=\frac{1}{3}\log \left(e^{3x+4}\right)+\log \left(e^{6x+8}\right)\]
A rule of logs:\[\large\rm \log\left(a^{\color{orangered}{b}}\right)=\color{orangered}{b}\log(a)\]Applying this rule gives us:\[\Large\rm y=\frac{1}{3}(3x+4)\log (e)+(6x+8)\log (e)\]

Answer 9

Then Find its derivative now ?

Answer 10

Yes.
Log base 10 of e is just a CONSTANT.
Don't let it confuse you!
We don't have to do any fancy product rule or anything like that.

Answer 11

\[\Large\rm y=\frac{\log(e)}{3}(3x+4)+\log(e)(6x+8)\]It's in the form:\[\Large\rm y=A(3x+4)+B(6x+8)\]Where A and B are constants.
So just differentiate what is inside of the brackets.

Answer 12

The after differentiating the thing inside the parenthesis . whats next that I need to do ?

Answer 13

\[\Large\rm y'=\frac{\log(e)}{3}(3)+(6)\log(e)\]Mmm nothing really :\
Just simplify the 3's.
I doubt this is the correct form of the question though.
It's really confusing, not knowing what they're actually asking.

Answer 14

Is the First solution which is The answer 1 / (3x+4) + 6 / (6x+8) . Is correct base on the BASE e or not ?

Answer 15

Yes, our first solution corresponded to an equation containing logs of base e.

Answer 16

\[y' =\frac{ 1 }{ 3x+4 }+ \frac{ 6 }{ 6x+8}\]

Answer 17

Is that Correct ?

Answer 18

Yes

Answer 19

So , Mr. Zepdrix What is the Answer on the second situation ?

Im sorry im giving you a hard time on those things that makes my mind confuse .
But Last week My prof . Give a unsure Answer . Which is 7loge I dont know if its correct or not . My prof also did not know If its correct.

Answer 20

For the second form, we assumed the logs were of base 10 and
applied a rule of logs before differentiating.
Then after taking our derivative, and simplifying, we end up with:\[\Large\rm y'=\log_{10}(e)+6\log_{10}(e)\]

Answer 21

Its also 7log e Right with Base of 10 ?

Answer 22

Oh yes yes. Like-terms, we can combine! :)
Correct.

Answer 23

Thank you so much
Btw im taking Diff Cal. now in school Thats Why I ask you about this sort of Questions.

Answer 24

Oh cool c:
Calculus is so much fun.

Answer 25

Do you have any website that will help me improve in Math ?

Answer 26

Khan Academy is a great resource.
https://www.khanacademy.org

I used this website a ton before taking Calculus.
https://www.math.ucdavis.edu/~kouba/ProblemsList.html
It really helped me to understand things and get ready for the subject.

MIT has their OpenCourseWare on Youtube, it's really excellent.
https://www.youtube.com/watch?v=7K1sB05pE0A&list=PL590CCC2BC5AF3BC1
https://www.youtube.com/watch?v=PxCxlsl_YwY&list=PL4C4C8A7D06566F38
https://www.youtube.com/watch?v=XDhJ8lVGbl8&list=PLEC88901EBADDD980

Answer 27

Wow Great!! Thank you so much