## Jaweria Group Title 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. one year ago one year ago

1. Curry

just post, i got you.

2. Jaweria

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

3. Jaweria

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

4. binarymimic

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

5. Jaweria

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

6. binarymimic

likewise for the second

7. Jaweria

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

8. binarymimic

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

9. Jaweria

ohh thank you so much

10. binarymimic

|dw:1359266294450:dw|

11. binarymimic

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

12. Jaweria

yeah i can

13. Jaweria

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

14. binarymimic

multiplication

15. Jaweria

ok

16. Jaweria

but then we have to simplify more right?

17. binarymimic

yeah you would have to isolate y'

18. Jaweria

ok can we go through this together till the end?

19. binarymimic

sure

20. Jaweria

thank u :)

21. binarymimic

|dw:1359266644304:dw|

22. binarymimic

|dw:1359266753423:dw|

23. Jaweria

how did u get -2?

24. binarymimic

oh my bad, that should be a 12

25. binarymimic

|dw:1359266876031:dw|

26. Jaweria

oh ok

27. binarymimic

|dw:1359266992506:dw|

28. Jaweria

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

29. binarymimic

where?

30. Jaweria

its ok I got it

31. binarymimic

ok. try the second one whenever you are comfortable

32. Jaweria

ok

33. Jaweria

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

34. Jaweria

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

35. binarymimic

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. Jaweria

ok

37. binarymimic

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. binarymimic

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. Jaweria

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. Jaweria

I m still learning

41. binarymimic

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. Jaweria

2x cos(x)

43. binarymimic

are you sure?

44. Jaweria

not much

45. binarymimic

don't forget the product rule

46. Jaweria

i always mess up on product rule and chain rule

47. binarymimic

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

48. binarymimic

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

49. Jaweria

ohh ok

50. binarymimic

now try differentiating y^2cos(y)

51. Jaweria

ok

52. Jaweria

not sure if I did it right or not

53. Jaweria

|dw:1359268366069:dw|

54. binarymimic

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. Jaweria

oh ok

56. binarymimic

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

57. Jaweria

yes

58. binarymimic

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

59. Jaweria

so that will be |dw:1359268690019:dw|

60. binarymimic

|dw:1359268763927:dw|

61. Jaweria

ok

62. binarymimic

you need to isolate dy/dx now

63. Jaweria

yeah ok

64. Jaweria

|dw:1359269043320:dw|

65. Jaweria

is that right?

66. binarymimic

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. Jaweria

ok thanks

68. binarymimic

wait let me double check something

69. Jaweria

couple more please if you dont mind?

70. Jaweria

ok

71. binarymimic

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

72. Jaweria

oh ok

73. binarymimic

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

74. Jaweria

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

75. Jaweria

Yeah I am trying my best.