## 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. 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. anonymous

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

3. AravindG

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

4. anonymous

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

5. anonymous

thank you :)

6. anonymous

and we are finding dx/dy

7. AravindG

ok diff both sides with respect to y

8. AravindG

use product rule

9. anonymous

ok

10. anonymous

Like this:|dw:1359270336745:dw|

11. anonymous

not sure though

12. anonymous

can anyone help me here with my questions?

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

ahan ok

17. AravindG

sorry i gt disconnected :O

18. anonymous

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

19. anonymous

oh its ok AravindG

20. anonymous

ok i will tell you

21. AravindG

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

22. anonymous

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

23. anonymous

everybody left me here :(

24. anonymous

no one wants to help me :(

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

Do you know how to differentiate explicitly?

29. anonymous

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

30. anonymous

$x^-2$

31. anonymous

oh yeah i do sorry

32. anonymous

$x^{-2}$

33. anonymous

So now, do you know how to differentiate explicitly?

34. anonymous

That means you differentiate boths sides at the same time.

35. anonymous

yup

36. anonymous

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. anonymous

ahan ok

38. anonymous

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

39. anonymous

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

40. anonymous

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

41. anonymous

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

42. anonymous

Then what's your next step.

43. anonymous

?*

44. anonymous

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

45. anonymous

oppss sorry i did bad

46. anonymous

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

47. anonymous

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

48. anonymous

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

49. anonymous

yes

50. anonymous

Why did you get y?

51. anonymous

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

52. anonymous

why did you get -x?

53. anonymous

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

54. anonymous

differentiate x^2

55. anonymous

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

56. anonymous

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

57. anonymous

Make dy/dx the subject.

58. anonymous

sorry one question that why there is

59. anonymous

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

60. anonymous

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

61. anonymous
62. anonymous

thanks Nitz for this :)

63. anonymous

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

64. anonymous

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

65. anonymous

yes I do

66. anonymous

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. anonymous

oh ok

68. anonymous

nitz just differentiate straight away using the product rule.

69. anonymous

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. anonymous

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

71. anonymous

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

72. anonymous

I found it hard

73. anonymous

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

74. anonymous

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

75. anonymous

Nitz is wrong then...

76. anonymous

<Look at his answer.

77. anonymous

He did dx/dy. You want dy/dx

78. anonymous

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

79. anonymous

oh yeah

80. anonymous

...

81. anonymous

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

82. anonymous

here's how you get -y/x

83. anonymous

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

84. anonymous

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

85. anonymous

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

86. anonymous

yup

87. anonymous

Then you make dy/dx as the subject.

88. anonymous

And you're done.

89. anonymous

ahan cool thanks.

90. anonymous

No worries mate.

91. anonymous

one last question plzz

92. anonymous

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

93. anonymous

alright.

94. anonymous

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

95. AravindG

to find dx/dy ?

96. anonymous

yup

97. AravindG

product rule on left

98. AravindG

try using draw button i will check :)

99. anonymous

hmm ok

100. anonymous

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

101. AravindG

try ! try you will succeed ! :)

102. anonymous

alright :)

103. anonymous

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. anonymous

ok sure

107. anonymous

|dw:1359274789970:dw|

108. AravindG

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

109. anonymous

oh ok sorry

110. AravindG

and why do you have 2 equal signs ?

111. anonymous

|dw:1359274989073:dw|

112. anonymous

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. anonymous

yeah thats what I pot in paranthesis next to the X

115. anonymous

put

116. AravindG

you wrote dy/dx there?

117. anonymous

yes next to it

118. anonymous

|dw:1359275398794:dw|

119. AravindG

what was the question ?

120. AravindG

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

121. anonymous

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

122. AravindG

|dw:1359275616596:dw|

123. AravindG

this is what i get

124. anonymous

sorry what is equal to dy/dx?

125. AravindG

nothing !!

126. AravindG

we just diff with respect to dy the whole equation

127. anonymous

oh ok

128. AravindG

so why do you think dy/dx will appear?

129. anonymous

so the way i did is that wrong?

130. AravindG

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

131. anonymous

ahan ok

132. AravindG

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

133. anonymous

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. anonymous

i have an answer sheet though

137. anonymous

alright then i ll work on them :)

138. AravindG

ok whats the answer you have for this question ?

139. anonymous

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. anonymous

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. anonymous

nope its asking for dy/dx

144. AravindG

then my working is right !! :)

145. anonymous

ok :)

146. AravindG

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

147. AravindG

understood?

148. anonymous

ok

149. anonymous

yup

150. anonymous

thanks :)

151. AravindG

Ok bye I gtg

152. anonymous

ok bye

153. AravindG

yw :)