## 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 Group Title

just post, i got you.

2. Jaweria Group Title

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

3. Jaweria Group Title

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

4. binarymimic Group Title

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

5. Jaweria Group Title

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

6. binarymimic Group Title

likewise for the second

7. Jaweria Group Title

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

8. binarymimic Group Title

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

9. Jaweria Group Title

ohh thank you so much

10. binarymimic Group Title

|dw:1359266294450:dw|

11. binarymimic Group Title

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

12. Jaweria Group Title

yeah i can

13. Jaweria Group Title

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

14. binarymimic Group Title

multiplication

15. Jaweria Group Title

ok

16. Jaweria Group Title

but then we have to simplify more right?

17. binarymimic Group Title

yeah you would have to isolate y'

18. Jaweria Group Title

ok can we go through this together till the end?

19. binarymimic Group Title

sure

20. Jaweria Group Title

thank u :)

21. binarymimic Group Title

|dw:1359266644304:dw|

22. binarymimic Group Title

|dw:1359266753423:dw|

23. Jaweria Group Title

how did u get -2?

24. binarymimic Group Title

oh my bad, that should be a 12

25. binarymimic Group Title

|dw:1359266876031:dw|

26. Jaweria Group Title

oh ok

27. binarymimic Group Title

|dw:1359266992506:dw|

28. Jaweria Group Title

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

29. binarymimic Group Title

where?

30. Jaweria Group Title

its ok I got it

31. binarymimic Group Title

ok. try the second one whenever you are comfortable

32. Jaweria Group Title

ok

33. Jaweria Group Title

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

34. Jaweria Group Title

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

35. binarymimic Group Title

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 Group Title

ok

37. binarymimic Group Title

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 Group Title

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 Group Title

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 Group Title

I m still learning

41. binarymimic Group Title

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 Group Title

2x cos(x)

43. binarymimic Group Title

are you sure?

44. Jaweria Group Title

not much

45. binarymimic Group Title

don't forget the product rule

46. Jaweria Group Title

i always mess up on product rule and chain rule

47. binarymimic Group Title

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

48. binarymimic Group Title

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

49. Jaweria Group Title

ohh ok

50. binarymimic Group Title

now try differentiating y^2cos(y)

51. Jaweria Group Title

ok

52. Jaweria Group Title

not sure if I did it right or not

53. Jaweria Group Title

|dw:1359268366069:dw|

54. binarymimic Group Title

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 Group Title

oh ok

56. binarymimic Group Title

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

57. Jaweria Group Title

yes

58. binarymimic Group Title

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

59. Jaweria Group Title

so that will be |dw:1359268690019:dw|

60. binarymimic Group Title

|dw:1359268763927:dw|

61. Jaweria Group Title

ok

62. binarymimic Group Title

you need to isolate dy/dx now

63. Jaweria Group Title

yeah ok

64. Jaweria Group Title

|dw:1359269043320:dw|

65. Jaweria Group Title

is that right?

66. binarymimic Group Title

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 Group Title

ok thanks

68. binarymimic Group Title

wait let me double check something

69. Jaweria Group Title

couple more please if you dont mind?

70. Jaweria Group Title

ok

71. binarymimic Group Title

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

72. Jaweria Group Title

oh ok

73. binarymimic Group Title

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

74. Jaweria Group Title

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

75. Jaweria Group Title

Yeah I am trying my best.