## anonymous 3 years ago I need help with 11 questions of Derivatives problems in Calculus 1. Can anyone help me with that please let me know, I will really appreciate it.

1. Curry

just post, i got you.

2. anonymous

ok first one is: (2x-3)^2+(4y-5)^2=10

3. anonymous

and we need to find dy/dx for all of them.

4. anonymous

you can either use implicit differentiation from the start, or you can begin by isolating the y variable.

5. anonymous

second question is: x^2sin(x)+y^2cos(y)=1

6. anonymous

likewise for the second

7. anonymous

yeah we have to do implicit diffentiation but I am really stuck in all of them.

8. anonymous

i'll get you started with (2x-3)^2+(4y-5)^2=10

9. anonymous

ohh thank you so much

10. anonymous

|dw:1359266294450:dw|

11. anonymous

idk if you can read the edge of it, it should say = 0

12. anonymous

yeah i can

13. anonymous

but after the paranthesis is that a multiply by 2 or its a power of 2 for the first one

14. anonymous

multiplication

15. anonymous

ok

16. anonymous

but then we have to simplify more right?

17. anonymous

yeah you would have to isolate y'

18. anonymous

ok can we go through this together till the end?

19. anonymous

sure

20. anonymous

thank u :)

21. anonymous

|dw:1359266644304:dw|

22. anonymous

|dw:1359266753423:dw|

23. anonymous

how did u get -2?

24. anonymous

oh my bad, that should be a 12

25. anonymous

|dw:1359266876031:dw|

26. anonymous

oh ok

27. anonymous

|dw:1359266992506:dw|

28. anonymous

and do we have to put y^1 before paranthesis?

29. anonymous

where?

30. anonymous

its ok I got it

31. anonymous

ok. try the second one whenever you are comfortable

32. anonymous

ok

33. anonymous

ok I have a feeling that I ll again get stuck with this one:

34. anonymous

x^2sin(x)+y^2cos(y)=1

35. anonymous

just think of differentiating y like you were differentiating any other variable (except x). remember to write the "dy/dx" or " y' " after you differentiate

36. anonymous

ok

37. anonymous

like here's an example: if i differentiate y*sin y with respect to x: = y cos(y) * y' + sin(y) * y' = y' [y cos(y) + sin(y)]

38. anonymous

it's the same as if you were differentiating 'u', as in differentiate: sin(x^2) let u = x^2 and differentiate sin(u) d/dx sin(u) = cos(u) * du = cos(x^2) * du Another way of saying that is: cos(x^2) * u' The only difference here is, we can write du in terms of x very easily. With implicit differentiation, we normally can't do that, but it's the same concept.

39. anonymous

hmmm if you dont mind can we go over this problem because my brain is not working at all with all of these i am sorry but thats the truth all my 11 questions are due tomorrow online :(

40. anonymous

I m still learning

41. anonymous

x^2sin(x)+y^2cos(y)=1 let's start with the easy part. if we're going to differentiate both sides, the right side becomes 0. what does x^2 sin(x) become ?

42. anonymous

2x cos(x)

43. anonymous

are you sure?

44. anonymous

not much

45. anonymous

don't forget the product rule

46. anonymous

i always mess up on product rule and chain rule

47. anonymous

$\frac{ d }{ dx } x^{2} \sin x = x^{2} \cos x + \sin x * 2x$

48. anonymous

"first times the derivative of the second, plus second times the derivative of the first"

49. anonymous

ohh ok

50. anonymous

now try differentiating y^2cos(y)

51. anonymous

ok

52. anonymous

not sure if I did it right or not

53. anonymous

|dw:1359268366069:dw|

54. anonymous

the left part is almost right, just need the negative sign (the derivative of cos is -sin the other part would become: cos(y) * 2y * dy/dx

55. anonymous

oh ok

56. anonymous

so i think you got that right, too, just needed the dy/dx ?

57. anonymous

yes

58. anonymous

$x^{2} \cos x + 2x \sin x + \frac{ dy }{ dx } * -y^{2} \sin y + \frac{ dy }{ dx } * 2y \cos y$

59. anonymous

so that will be |dw:1359268690019:dw|

60. anonymous

|dw:1359268763927:dw|

61. anonymous

ok

62. anonymous

you need to isolate dy/dx now

63. anonymous

yeah ok

64. anonymous

|dw:1359269043320:dw|

65. anonymous

is that right?

66. anonymous

yep perfect you can eliminate the minus sign on the outside if you wanted to by swapping the terms in the denominator to make it: y^2 sin y - 2y cos y but thats just a matter of preference

67. anonymous

ok thanks

68. anonymous

wait let me double check something

69. anonymous

couple more please if you dont mind?

70. anonymous

ok

71. anonymous

ok. sorry. if you had more questions you might want to start a new post

72. anonymous

oh ok

73. anonymous

after a few of these, implicit differentiation will be easy for you

74. anonymous

this is the first time ever I am using this site can you suggest me that how to start new post?

75. anonymous

Yeah I am trying my best.