## anonymous 4 years ago solve the derivatives (e^3x)/(4x+1) so far i got $(e ^{3x}(12x-1))/(16x ^{2}+4x+1$

1. amistre64

solve the derivatives? or calculate the derivatives?

2. amistre64

$e^{3x}(4x+1)^{-1}$ $e'^{3x}(4x+1)^{-1}+ e^{3x}(4x+1)'^{-1}$

3. anonymous

calculate sorry

4. amistre64

$e'^{3x}(4x+1)^{-1}+ e^{3x}(4x+1)'^{-1}$ $3e^{3x}(4x+1)^{-1} -4e^{3x}(4x+1)^{-2}$should be the first derivative

5. amistre64

it tends to be simpler to do product rule than quotient rule

6. anonymous

im really sorry but i dont seem to follow what you did

7. amistre64

$(3e^{3x}(4x+1) -4e^{3x})(4x+1)^{-2}$ $\frac{3e^{3x}(4x+1) -4e^{3x}}{(4x+1)^{2}}$ can be reconstructed

8. amistre64

theres an inverse rule $\frac{1}{a}=a^{-1}$

9. amistre64

i just converted the denominator up into an exponent of -1 to use the product rule on it

10. anonymous

ya i multiplied the 3e^xx to the 4x and the 1 and then combined like terms and ultimatly removing a e^3x

11. amistre64

if you need more derivatives; its best to keep it in product form

12. amistre64

yeah, so far you did fine

13. amistre64

how many derivatives you need to take it to? or is the one good enough?

14. anonymous

ya its just one i think it says find f'(x)

15. anonymous

the ' indicating 1 derivative correct?

16. amistre64

then yeah, this is fine; just simplify it as far as you like or leave it as is. $\frac{3e^{3x}(4x+1) -4e^{3x}}{(4x+1)^{2}}$ as long as you have places to plug in your "x" values to determines slopes thats fine

17. anonymous

really?

18. amistre64

yes, f' means 1st derivative :)

19. amistre64

yeah, prettying it up doesnt change its value one bit

20. anonymous

i dont have to simplify it down?

21. amistre64

not to use it, no

22. amistre64

making something pretty has no consequance on the values it produces

23. amistre64

3x + 4x produces the same results for and x value as 7x does right?

24. anonymous

you dont think the professor will mark it down

25. amistre64

dunno, i aint got your proffessor :)

26. anonymous

lol good point well thank you. i must admit i did well for just learning this stuff today

27. amistre64

yeah, keep it up :)

28. anonymous

even though using this in a physics 100 course seems a bit much

29. amistre64

derivatives are use din physics to model certain natural phenomenon and such; its good to know the concepts; but computers and such do the real work these days

30. anonymous

ya but teaching this in a course

31. anonymous

that is has a co-requisite with calc seems a bit rough, i m in calc im only learning about limists

32. amistre64

yeah, limits. something touted about but never really seen again

33. anonymous

good because they drive my crazy

34. amistre64

35. anonymous

its like im just going to be as accurate as i want to be because there is an infinite space between points

36. amistre64

well, good luck :)

37. anonymous

have a good one and thanks again