## anonymous 3 years ago HELP Please. Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=4 at the point (2,1) . The equation of this tangent line can be written in the form: y=mx+b where M is: and where B is:

1. anonymous

remember product rule for implicit $(fg)' = f'g +fg'$

2. anonymous

M = slope = dy/dx

3. anonymous

oh so for implicit we use the product?

4. anonymous

always?

5. anonymous

yes and chain rule....since it is with respect to x $d/dx f(y) = f'(y)*\frac{dy}{dx}$

6. anonymous

so which is f and which is g

7. anonymous

im a little confused since xy^3 are together.

8. anonymous

depends on the term...first there is "xy^3" f(x) = x g(y) = y^3

9. anonymous

so f = x and g= y^3 ?

10. anonymous

so is the derivative: x*3y+ y^3*1 ?

11. anonymous

yes, so result is $1*y^{3}+ x*3y^{2} \frac{dy}{dx}$

12. anonymous

dy/dx?

13. anonymous

thats what the slope is...which is what we are trying to find

14. anonymous

all i know is the product rule is this: $\frac{ d }{ dx } (fg)= f g \prime+g f \prime$

15. anonymous

anytime you differentiate a function of "y" wrt x you must multiply by "dy/dx" refer to my prev post

16. anonymous

otay. so then what.

17. anonymous

repeat process for all terms of equation

18. anonymous

?

19. anonymous

what do you mean

20. anonymous

xy^3 +xy = 4 you have to differentiate each term, the "xy^3" and "xy" adn "4"

21. anonymous

3xy^2.....

22. anonymous

?? first term is $(y^{3} +3xy^{2} \frac{dy}{dx})$

23. anonymous

next term is "xy" .... derivative is $(y + x \frac{dy}{dx})$ then of course , derivative of constant is 0 now you have $y^{3} +3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0$

24. anonymous

solve for "dy/dx" in terms of x and y

25. anonymous

okay i understand from here: $\frac{ dy }{ dx}= 3xy ^{2}+ y ^{3}$

26. anonymous

then from there, i don't know what to do.

27. anonymous

@dumbcow

28. anonymous

well from there you would plug in the given point (2,1) to give you the slope (M) of tangent line but how did you solve for dy/dx .... your answer is wrong

29. anonymous

i didn't i'm doing this step by step

30. anonymous

from the result.. ^^ i just multiplied.

31. anonymous

?? sorry i am not understanding

32. anonymous

remember before you wrote "yes, result is bla blah" then from there i just multiplied... x*(3y)^2 + y^3 *1

33. anonymous

you know.

34. dan815

hey mathcalculus

35. dan815

do you know chain rule

36. dan815

|dw:1363334172119:dw|

37. anonymous

let me just show the steps, maybe that will help $y^{3}+3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0$ $3xy^{2} \frac{dy}{dx}+x \frac{dy}{dx} = -y^{3}-y$ $\frac{dy}{dx}(3xy^{2}+x) = -y^{3}-y$ $\frac{dy}{dx} = \frac{-y^{3}-y}{3xy^{2}+x}$

38. anonymous

i get the feeling i lost you when i started using "dy/dx" :{

39. anonymous

mhm :/

40. anonymous

|dw:1363334405957:dw| @dan815

41. dan815

|dw:1363334256950:dw|

42. dan815

think about it like this Y is a function of X right so say you have to take the derivative of 2x^2 what is it?

43. dan815

|dw:1363334523512:dw|

44. anonymous

wait wait wait

45. anonymous

thx @dan815 for clarifying :) i didn't explain very well

46. dan815

remember how chain rule works

47. anonymous

the chain rule: if f(x)= g(k(x)) then f prime x= g prime (k(x))k prime (x) correct?

48. anonymous

@dumbcow you explained very good too.

49. dan815

@dumbcow no its okay lol i can relate to him cuz i know how noobs think :)

50. anonymous

i just can't understand @dan815 techwriting lol

51. dan815

yes that is chain rule

52. anonymous

noob? -_-

53. anonymous

411- not a guy.

54. dan815

haha trust me we are all noobs, just wait till you see how much those phd profs know

55. anonymous

okay dc back to math.

56. anonymous

thank god, no more crazy math after this. but seriously back to work lol

57. anonymous

i have to understand this.

58. dan815

ok ok so for implicit differentiation when you see Y think about it like some function on x

59. anonymous

ello?

60. anonymous

yup got that.

61. anonymous

then?

62. dan815

ok do this

63. anonymous

so first i was told to use the product rule.. then the chain. which that confused me a little. so lets take it step by step please.

64. dan815

|dw:1363334909090:dw|

65. dan815

yes you are going to need both those rules for this

66. dan815

no remember than y is like a function of x

67. anonymous

OH

68. anonymous

darn it.

69. dan815

|dw:1363334981033:dw|

70. anonymous

so we use the product rule 1st.

71. dan815

sub 2x where you see Y and solve it

72. anonymous

i have no idea what im doing.

73. anonymous

for some reason I'm all confused.

74. dan815

okay look at this question again same question but im telling you what y is

75. anonymous

don't tell me what y is. how am i suppose to know alone?

76. anonymous

let's do that problem that I gave.

77. dan815

|dw:1363335163468:dw|

78. dan815

just do this im trying to teach you how to do the problem above and every other problem like this

79. anonymous

k -.-

80. anonymous

x= 2/5 ?

81. anonymous

i plugged in 2x..

82. dan815

wut no find dervivative

83. dan815

chain rule

84. anonymous

?

85. dan815

|dw:1363335349192:dw|

86. dan815

do you get how i got that when i differentiated there

87. dan815

whats confusing you there is it the d/dx?

88. dan815

i know that you know how to take derivate of Xs

89. dan815

you there?

90. anonymous

91. anonymous

yeah im looking.

92. dan815

ya i think so lol we've found the root problem here

93. dan815

she just doesnt know what it means to difference functions as opposed to variables

94. anonymous

@dumbcow you used the product rule first which i was understanding.... until mr. CHAIN RULE came along to make a knot.

95. anonymous

:)

96. dan815

|dw:1363335696884:dw|

97. dan815

do you understand this yet?

98. anonymous

yeah...sorry i figured that was implied since we are dealing with implicit differentiation :|

99. anonymous

yeah @dan815 but now why did you multiply 2 at the end.?

100. dan815

thats chain rule remember you have to multiply by the derivative of the function inside the bracket

101. anonymous

i understand you got the derivative

102. anonymous

|dw:1363335825708:dw|

103. dan815

|dw:1363335821148:dw|

104. dan815

this is known as the chain rule

105. anonymous

okie let's do our problem @dan815

106. anonymous

i know youre trying to teach me, so i can do the rest.. etc; but i need to do one problem first in order to proceed.

107. anonymous

i remember the chain rule a little better now.

108. dan815

so if its |dw:1363335898535:dw|

109. dan815

ok lets go back to our problem knowing the chain rule now

110. anonymous

gratzie

111. dan815

what i really wanted you to undetsand was this concept though basically you have to think of Y as a function so when you see Ys differentiating Y is like differentiation of a function so think chain rules

112. dan815

|dw:1363336027415:dw|

113. dan815

so for the implict up there

114. anonymous

ok

115. anonymous

got it, think the x as in y.

116. anonymous

so our problem... is this: xy^3+xy=4 at the point (2,1) .

117. anonymous

start by:

118. dan815

|dw:1363336084849:dw|

119. dan815

|dw:1363336247542:dw|