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## KarateChopKid Group Title How would one differentiate the following: 2x+3/√(1-2x) 2 years ago 2 years ago

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1. baddinlol Group Title

You can use the quotient rule.

2. baddinlol Group Title

Are you familiar with that?

3. KarateChopKid Group Title

yes I am familiar with this

4. baddinlol Group Title

so u = 2x+3 u' = 2 v = √(1-2x) v' = -1/√(1-2x)

5. baddinlol Group Title

The quotient rule says dy/dx = (u'v - uv')/v^2

6. KarateChopKid Group Title

ok and from there I begin to distribute, correct?

7. gurvinder Group Title

u can do it with quotient rule

8. baddinlol Group Title

Thats it follow (u'v - uv')/v^2

9. shana91 Group Title

|dw:1331119146075:dw| use this this is the division rule

10. shana91 Group Title

where u and v are two fuctions

11. KarateChopKid Group Title

I think that the square root in the function os throwing me off but I think that it would disappear in the deno. right? just leaving 1-2x on the bottom?

12. shana91 Group Title

exactly. you are right. square root dissapears. THat makes the sum a lot easier

13. .Sam. Group Title

14. KarateChopKid Group Title

Thank you! I was working it out and ended with the same result.

15. KarateChopKid Group Title

One final question... how would you find the eq. of a line tangent to the graph f(x) = √(2x+9) at x=0 ?

16. shana91 Group Title

take the first derivative of f(x) and equate all x to zero it gives the gradient of the tangent to the line at x=0

17. .Sam. Group Title

differentiate √(2x+9) then equate x=0 to the differentiated expression and get your gradient then create new equation using y-y1=m(x-x1)

18. shana91 Group Title

then use y=mx+c to get the equa. of the line

19. .Sam. Group Title

to find y1 just equate x=0 to original equation, y=√(2x+9), then get your y1

20. KarateChopKid Group Title

ok I am still a little shaky on deffrentiating a radical equation, would it be 1/2x+9 ?

21. .Sam. Group Title

|dw:1331120584519:dw|

22. shana91 Group Title

when u dof. the above one it'll be 1/(2x+9)^1/2

23. shana91 Group Title

you have forgot the 1/2

24. .Sam. Group Title

1/2 cancelled by 2

25. .Sam. Group Title

chain rule

26. KarateChopKid Group Title

ok so using the chain rule I use a fraction to symbolize the same radical function then move the 1/2 down and subtract by 1 to get my new power...got it

27. KarateChopKid Group Title

where did the "2" on the right come from though?

28. shana91 Group Title

not that 1/2 i meant the power 1/2|dw:1331120933568:dw| this should be what u getting aftr dif.

29. .Sam. Group Title

Example, |dw:1331120907510:dw|

30. KarateChopKid Group Title

Ok, so since it was in the deno. it was risen to the 2nd power and that is how is came to cancel the 1/2 right?

31. KarateChopKid Group Title

in the orig. differentation

32. .Sam. Group Title

the 2 in the numerator when differentiating, so it will cancel the 1/2

33. .Sam. Group Title

$\huge \sqrt{x}=x ^{\frac{1}{2}}$

34. KarateChopKid Group Title

ok I undrstand that example in omitting the radical but where did the 2 appear to make that cancel?

35. KarateChopKid Group Title

that is what is throwing me for a loop haha

36. .Sam. Group Title

the half of 2 is 1

37. shana91 Group Title

wait --look at this.. this how u solve n this is whr it got cancelled|dw:1331121592002:dw|

38. KarateChopKid Group Title

nevermind it came as a result of the y' and omitting the 9

39. KarateChopKid Group Title

or making 2x+9 just 2 right?

40. shana91 Group Title

I cant get u?

41. KarateChopKid Group Title

lol I can't get myself sometime haha but I think I understand that the 2 came as a result of taking the derivative of just 2x+9 resulting in just the 2 remaining...I think : /

42. shana91 Group Title

ohh that one..wait this is clear the prob.|dw:1331122282837:dw| cool?

43. KarateChopKid Group Title

yea... that I what I was trying to say but I guess my wording is off. I see where the 2 came from

44. shana91 Group Title

:)

45. KarateChopKid Group Title

Haha thanks. So then from there I plug in my x=0 to solve for that equation

46. KarateChopKid Group Title

and then use that solution in point-slope form

47. shana91 Group Title

that is correct. :) and then use thegradient equation to find the gradient...and the eq. of the line

48. salini Group Title

hey shana as u say the 2 in the denominator cannot get cancelled...

49. KarateChopKid Group Title

which gives me 3 from the graident and 1/3 for my derivative function

50. KarateChopKid Group Title

|dw:1331123039074:dw|

51. KarateChopKid Group Title

|dw:1331123158889:dw|

52. KarateChopKid Group Title

correct?

53. shana91 Group Title

correct

54. KarateChopKid Group Title

good deal! Thank you for all of you help and have an awesome day or night wherever you are!

55. KarateChopKid Group Title

|dw:1331123407516:dw|

56. shana91 Group Title

hhe it'll be night :) Anytime . I love guiding others not for money but for joy. :)

57. salini Group Title

again how di u agree abt that 2 getting cancelled in the differentaition part

58. shana91 Group Title

well salini, when u intergrate, the power comes to the front. ok? but when u again use the chain rule to integrate 2X u have to again multiply by 2. so they get cancelled off.

59. salini Group Title

oh yeah 2x part......

60. salini Group Title

we cant give u money here instead we give away medals!

61. shana91 Group Title

I really dont do this for money.... your medals will be highly appreciated.. I just joined today n already in level 10. :) This is a good place to give away my knowledge.