## Jaweria 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. AravindG

we cant help with all 11 .maybe we can help in 1 or 2 with fond hope that you will understand how to do these and finish the rest yourself.

2. Jaweria

hmm thanks ok i ll need your help with atleast 2 of them.

3. AravindG

bring it on !! and "WELCOME TO OPENSTUDY!!"

4. Jaweria

one of them is : |dw:1359270057247:dw|

5. Jaweria

thank you :)

6. Jaweria

and we are finding dx/dy

7. AravindG

ok diff both sides with respect to y

8. AravindG

use product rule

9. Jaweria

ok

10. Jaweria

Like this:|dw:1359270336745:dw|

11. Jaweria

not sure though

12. Jaweria

can anyone help me here with my questions?

13. nitz

its $x ^{2}y ^{2}=1$ or $xy ^{2}=1$?

14. Jaweria

its x$its x ^{2}y ^{2}=1$

15. nitz

OK....differentiate both sides wrt to y using product rule:: $x ^{2}(2y)+((y ^{2})2x (dx/dy))=0$

16. Jaweria

ahan ok

17. AravindG

sorry i gt disconnected :O

18. nitz

so tell me from above equation what would be dx/dy?

19. Jaweria

oh its ok AravindG

20. Jaweria

ok i will tell you

21. AravindG

just solve for dx/dy after you applied product rule which will be .....

22. Jaweria

I need little bit of more discription for that if its possible.

23. Jaweria

everybody left me here :(

24. Jaweria

no one wants to help me :(

25. Azteck

$y^2=\frac{ 1 }{ x^2 }$

26. Azteck

$y^2=x^{-2}$

27. Jaweria

how did you get $X ^{-2}$

28. Azteck

Do you know how to differentiate explicitly?

29. Azteck

You don't know that ???? $x^-2=\frac{1}{x^2}$

30. Azteck

$x^-2$

31. Jaweria

oh yeah i do sorry

32. Azteck

$x^{-2}$

33. Azteck

So now, do you know how to differentiate explicitly?

34. Azteck

That means you differentiate boths sides at the same time.

35. Jaweria

yup

36. Azteck

Okay. Give it a shot and tell me what you get. Remember, when you differentiate the y part, you also have to attach a dy/dx.

37. Jaweria

ahan ok

38. Jaweria

is it $-\frac{ y }{ x}$

39. Azteck

Nope. Show me what you did as your first step.

40. Azteck

I gave you this to differentiate. $y^2=x^{-2}$

41. Jaweria

I did like this:|dw:1359271872658:dw|

42. Azteck

Then what's your next step.

43. Azteck

?*

44. Jaweria

then I canceled $Y ^{2} which became Y and Then I can canceled X ^{-2} and \it became -x$

45. Jaweria

oppss sorry i did bad

46. nitz

u have any doubt in my way......should i explain?

47. Jaweria

lolz right now I dont have any doubt at anyone I just need help very badly :(

48. Azteck

When you differentiate y^2, you get 2y?

49. Jaweria

yes

50. Azteck

Why did you get y?

51. Azteck

and when you differentiate x^{-2}, you get -2x^{-3}

52. Azteck

why did you get -x?

53. Jaweria

Yes I did a mistake for -X though but I couldnt understand Y one

54. Azteck

differentiate x^2

55. Azteck

That's exactly the same as differentiating y^2. What are you doing?

56. Azteck

$2y\times \frac{ dy }{ dx }=-\frac{ 2 }{ x^3 }$

57. Azteck

Make dy/dx the subject.

58. Jaweria

sorry one question that why there is

59. Jaweria

Sorry but i have a question that how you got X^3

60. nitz

$(2x ^{2}y)=-(2xy ^{2})(dx/dy)$ $dx/dy=-(2x ^{2}y)/(2xy ^{2})$ dx/dy=-x/y

61. Azteck
62. Jaweria

thanks Nitz for this :)

63. Jaweria

Aztec actually the answer I am having from the professor is matching with Nitz

64. Azteck

@Jaweria, do you have an answer sheet or something?

65. Jaweria

yes I do

66. Azteck

Ah okay. I can see that because you differentiate straight away. I moved the x^2 to the other side. Somehow that affected it.

67. Jaweria

oh ok

68. Azteck

nitz just differentiate straight away using the product rule.

69. Jaweria

but as Nitz just explained I am little bit confused there that how he is getting 2$2X ^{2}Y$ isnt that when you get 2X then you dont need $2X ^{2}$

70. Azteck

Don't you know how to do the product rule?

71. Jaweria

I have it on my notes and I am still trying to learn this is my first homework of this Calculus class

72. Jaweria

I found it hard

73. Azteck

What's the answer say in your sheet, Does it say -x/y or -y/x?

74. Jaweria

it says $-\frac{ y }{ x }$

75. Azteck

Nitz is wrong then...

76. Azteck

<Look at his answer.

77. Azteck

He did dx/dy. You want dy/dx

78. Jaweria

yeah but I dont think so that really matters you can do either way

79. Jaweria

oh yeah

80. Azteck

...

81. Jaweria

so its just we are flipping dy/dx instead of dx/dy

82. Azteck

here's how you get -y/x

83. Azteck

$u=x^2$ $u'=2x$ $v=y^2$ $v'=2y$ $\frac{ dy }{ dx }=(u'\times v)+(u\times v')$

84. Azteck

Sorry about that. $0=2xy^2+2x^2y(\frac{dy}{dx})$

85. Azteck

When you differentiate y^2 you would write dy/dx next to it.

86. Jaweria

yup

87. Azteck

Then you make dy/dx as the subject.

88. Azteck

And you're done.

89. Jaweria

ahan cool thanks.

90. Azteck

No worries mate.

91. Jaweria

one last question plzz

92. Azteck

Okay. last one. And you can try and finish the others yourself using what you got from here.

93. Jaweria

alright.

94. Jaweria

$y ^{2}\tan(x)=x$

95. AravindG

to find dx/dy ?

96. Jaweria

yup

97. AravindG

product rule on left

98. AravindG

try using draw button i will check :)

99. Jaweria

hmm ok

100. Jaweria

I know I ll get stuck again :( but I am trying

101. AravindG

try ! try you will succeed ! :)

102. Jaweria

alright :)

103. Jaweria

ok first step so far: |dw:1359274476783:dw|

104. AravindG

can you write thhird term again ? i cant see it clearly

105. AravindG

*third

106. Jaweria

ok sure

107. Jaweria

|dw:1359274789970:dw|

108. AravindG

i mean third term fully from y^2 sec^2 x

109. Jaweria

oh ok sorry

110. AravindG

and why do you have 2 equal signs ?

111. Jaweria

|dw:1359274989073:dw|

112. Jaweria

I put X after equal sign because thats also in the question

113. AravindG

for x diff with respect to y it will be simply dx/dy isnt it?

114. Jaweria

yeah thats what I pot in paranthesis next to the X

115. Jaweria

put

116. AravindG

you wrote dy/dx there?

117. Jaweria

yes next to it

118. Jaweria

|dw:1359275398794:dw|

119. AravindG

what was the question ?

120. AravindG

what i think is your question is $y^2 \tan(x)=x$

121. Jaweria

$Y ^{2}\tan (x)=x$

122. AravindG

|dw:1359275616596:dw|

123. AravindG

this is what i get

124. Jaweria

sorry what is equal to dy/dx?

125. AravindG

nothing !!

126. AravindG

we just diff with respect to dy the whole equation

127. Jaweria

oh ok

128. AravindG

so why do you think dy/dx will appear?

129. Jaweria

so the way i did is that wrong?

130. AravindG

if you wanted dx/dy I believe my method is enough ...

131. Jaweria

ahan ok

132. AravindG

i will be glad if someone verifies this bcoz i have not done differentiation for a while :P

133. Jaweria

is Aztec here?

134. AravindG

dunno , maybe he is away

135. AravindG

have patience :) in the meantime try to work on other questions .BEST OF LUCK !

136. Jaweria

i have an answer sheet though

137. Jaweria

alright then i ll work on them :)

138. AravindG

ok whats the answer you have for this question ?

139. Jaweria

its $(1-y ^{2}\sec ^{2}x)/2y \tan (x)$

140. AravindG

in btw i frgt to add tan x here |dw:1359276176708:dw|

141. Jaweria

ahan thats what I was thinking

142. AravindG

i get dx/dy as reciprocal of your answer...are you sure the question asks for dx/dy ?

143. Jaweria

nope its asking for dy/dx

144. AravindG

then my working is right !! :)

145. Jaweria

ok :)

146. AravindG

solve for dx/dy from my working...dy/dx=1/dx/dy

147. AravindG

understood?

148. Jaweria

ok

149. Jaweria

yup

150. Jaweria

thanks :)

151. AravindG

Ok bye I gtg

152. Jaweria

ok bye

153. AravindG

yw :)

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