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  # Differentiation question watch
  
  1. i did a differentiation question and i wanted to know if its the right answer.
  
  the q is: [(x-1)/(2-x)]^2 and i got [2(x-1)]/[(2-x)^3] sorry dont know how to use latex.
  
  i havent done diff for ages so im a bit rusty and wanted to know if its right and if not, where im going wrong.
  
  thanks much appreciated
  
  btw i used the chain rule and differentiated it as a whole and for the bit inside the brackets i used the quotient rule.
  2. I have got the same answer as you, I think you're right.
  3. (Original post by Hormonal)
  I have got the same answer as you, I think you're right.
  thanks for having a look
  4. i also have another question and i'd just like someone to check it for me.
  
  its: x=2cos(theta) and y=3sin(theta)
  
  find dy/dx in terms of theta, and the coordinates of the 2 turning points.
  
  Spoiler:
  Show
  I got dy/dx = 3cos(theta)/-2sin(theta)
  
  then at TP dy/dx = 0 so 3cos(theta) = 0.
  
  so theta must be 90 and/or 270.
  
  I subbed this back into the orig x and y and got the coordinates to be (0,3) and (0,-3).
  
  am I doing this right?
  
  thanks so much
  5. (Original post by SurpriseMe)
  i also have another question and i'd just like someone to check it for me.
  
  its: x=2cos(theta) and y=3sin(theta)
  
  find dy/dx in terms of theta, and the coordinates of the 2 turning points.
  
  Spoiler:
  Show
  I got dy/dx = 3cos(theta)/-2sin(theta)
  
  then at TP dy/dx = 0 so 3cos(theta) = 0.
  
  so theta must be 90 and/or 270.
  
  I subbed this back into the orig x and y and got the coordinates to be (0,3) and (0,-3).
  
  am I doing this right?
  
  thanks so much
  This is correct, though if I were being picky I'd put the angles in radians
  6. (Original post by joostan)
  This is correct, though if I were being picky I'd put the angles in radians
  lol, to be fair I did do it in radians cos my calc is always in that mode but then for some reason I just felt safer doing it in degrees cos i'm used to the angles at which the cos graph meets the x axis
  7. (Original post by SurpriseMe)
  lol, to be fair I did do it in radians cos my calc is always in that mode but then for some reason I just felt safer doing it in degrees cos i'm used to the angles at which the cos graph meets the x axis
  Meh, it's only a technicality, I know lots of people who do the same thing
  8. (Original post by joostan)
  Meh, it's only a technicality, I know lots of people who do the same thing
  yeah same here
  
  I have another question. How would you diff 2sin^3(e^x)?
  
  would you chnage it to 2[sin(e^x)]^3 and use the chain rule or something?
  9. bump.
  
  How would you diff 2sin^3(e^x)?
  
  would you change it to 2[sin(e^x)]^3 and use the chain rule or something?
  
  thanks
  10. bump
  11. (Original post by SurpriseMe)
  bump.
  
  How would you diff 2sin^3(e^x)?
  
  would you change it to 2[sin(e^x)]^3 and use the chain rule or something?
  
  thanks
  Product Rule:
  
  y = (u)(v)
  
  dy/dx = (u')(v) + (u)(v')
  
  Okay. Ignore me actually. Wow
  12. (Original post by SurpriseMe)
  bump.
  
  How would you diff 2sin^3(e^x)?
  
  would you change it to 2[sin(e^x)]^3 and use the chain rule or something?
  
  thanks
  Hey, yep it's just the chain rule with three steps:
  
  sin
  e^x
  [...]^3
  
  Multiply by the differential of all of those. Don't forget to include the 2 at the start as well!
  
  Hope that helps?
  
  Posted from TSR Mobile
  13. (Original post by L'Evil Fish)
  Product Rule:
  
  y = (u)(v)
  
  dy/dx = (u')(v) + (u)(v')
  Why product rule?
  
  (Original post by stuart_aitken)
  Hey, yep it's just the chain rule with three steps:
  
  sin
  e^x
  [...]^3
  
  Multiply by the differential of all of those. Don't forget to include the 2 at the start as well!
  
  Hope that helps?
  So it would be 6[sin(e^x)]^2 mutliplied by the differential of what's inside the brackets which is e^xcos(e^x)? Is my differential inside the bracket correct?
  14. (Original post by L'Evil Fish)
  Product Rule:
  
  y = (u)(v)
  
  dy/dx = (u')(v) + (u)(v')
  
  Okay. Ignore me actually. Wow
  What lol? I'm so confused
  
  you're probably way better at maths than i so.... :/
  15. (Original post by SurpriseMe)
  What lol? I'm so confused
  
  you're probably way better at maths than i so.... :/
  I thought we could use it like this:
  
  2Sin^3(e^x)
  
  But then I realised there wasn't an x after Sin^3.
  
  And yes, your differential outside the bracket is right
  16. (Original post by L'Evil Fish)
  I thought we could use it like this:
  
  2Sin^3(e^x)
  
  But then I realised there wasn't an x after Sin^3.
  
  And yes, your differential outside the bracket is right
  Ohhh yeah that would make sense
  
  thank you for checking as well
  17. (Original post by SurpriseMe)
  Ohhh yeah that would make sense
  
  thank you for checking as well
  No problem, my confidence has been knocked a little so I'm wary of answering but I do it anyway
  18. (Original post by L'Evil Fish)
  No problem, my confidence has been knocked a little so I'm wary of answering but I do it anyway
  haha, well as you can see my confidence isn't exactly tip-top otherwise I wouldn't keep asking how to do questions that I'm getting right
  
  well I don't blame you cos if you gave a wrong answer you might have a billion quotes questioning your abilities lol. That's why I stay far away from even answering gcse question
  19. (Original post by SurpriseMe)
  haha, well as you can see my confidence isn't exactly tip-top otherwise I wouldn't keep asking how to do questions that I'm getting right
  
  well I don't blame you cos if you gave a wrong answer you might have a billion quotes questioning your abilities lol. That's why I stay far away from even answering gcse question
  Fair enough
  
  Yeah, lately I've been getting a bit wrong so I haven't answered in a while
  
  What year are you in now?
  20. (Original post by L'Evil Fish)
  Fair enough
  
  Yeah, lately I've been getting a bit wrong so I haven't answered in a while
  
  What year are you in now?
  
  I'm in year 13, i'm supposed to be an A* student but i think i lack a bit of self-confidence sometimes and i hate being wrong lol :/
  
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explanation: The document contains a series of math questions and answers related to differentiation. The user initially asks for verification of their solution to a differentiation problem and then proceeds to ask for help with another differentiation problem. The community members provide feedback and confirm the correctness of the user's solutions.
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  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.content: You are Here: Home >< Maths
  
  # Differentiation question watch
  
  1. i did a differentiation question and i wanted to know if its the right answer.
  
  the q is: [(x-1)/(2-x)]^2 and i got [2(x-1)]/[(2-x)^3] sorry dont know how to use latex.
  
  i havent done diff for ages so im a bit rusty and wanted to know if its right and if not, where im going wrong.
  
  thanks much appreciated
  
  btw i used the chain rule and differentiated it as a whole and for the bit inside the brackets i used the quotient rule.
  2. I have got the same answer as you, I think you're right.
  3. (Original post by Hormonal)
  I have got the same answer as you, I think you're right.
  thanks for having a look
  4. i also have another question and i'd just like someone to check it for me.
  
  its: x=2cos(theta) and y=3sin(theta)
  
  find dy/dx in terms of theta, and the coordinates of the 2 turning points.
  
  Spoiler:
  Show
  I got dy/dx = 3cos(theta)/-2sin(theta)
  
  then at TP dy/dx = 0 so 3cos(theta) = 0.
  
  so theta must be 90 and/or 270.
  
  I subbed this back into the orig x and y and got the coordinates to be (0,3) and (0,-3).
  
  am I doing this right?
  
  thanks so much
  5. (Original post by SurpriseMe)
  i also have another question and i'd just like someone to check it for me.
  
  its: x=2cos(theta) and y=3sin(theta)
  
  find dy/dx in terms of theta, and the coordinates of the 2 turning points.
  
  Spoiler:
  Show
  I got dy/dx = 3cos(theta)/-2sin(theta)
  
  then at TP dy/dx = 0 so 3cos(theta) = 0.
  
  so theta must be 90 and/or 270.
  
  I subbed this back into the orig x and y and got the coordinates to be (0,3) and (0,-3).
  
  am I doing this right?
  
  thanks so much
  This is correct, though if I were being picky I'd put the angles in radians
  6. (Original post by joostan)
  This is correct, though if I were being picky I'd put the angles in radians
  lol, to be fair I did do it in radians cos my calc is always in that mode but then for some reason I just felt safer doing it in degrees cos i'm used to the angles at which the cos graph meets the x axis
  7. (Original post by SurpriseMe)
  lol, to be fair I did do it in radians cos my calc is always in that mode but then for some reason I just felt safer doing it in degrees cos i'm used to the angles at which the cos graph meets the x axis
  Meh, it's only a technicality, I know lots of people who do the same thing
  8. (Original post by joostan)
  Meh, it's only a technicality, I know lots of people who do the same thing
  yeah same here
  
  I have another question. How would you diff 2sin^3(e^x)?
  
  would you chnage it to 2[sin(e^x)]^3 and use the chain rule or something?
  9. bump.
  
  How would you diff 2sin^3(e^x)?
  
  would you change it to 2[sin(e^x)]^3 and use the chain rule or something?
  
  thanks
  10. bump
  11. (Original post by SurpriseMe)
  bump.
  
  How would you diff 2sin^3(e^x)?
  
  would you change it to 2[sin(e^x)]^3 and use the chain rule or something?
  
  thanks
  Product Rule:
  
  y = (u)(v)
  
  dy/dx = (u')(v) + (u)(v')
  
  Okay. Ignore me actually. Wow
  12. (Original post by SurpriseMe)
  bump.
  
  How would you diff 2sin^3(e^x)?
  
  would you change it to 2[sin(e^x)]^3 and use the chain rule or something?
  
  thanks
  Hey, yep it's just the chain rule with three steps:
  
  sin
  e^x
  [...]^3
  
  Multiply by the differential of all of those. Don't forget to include the 2 at the start as well!
  
  Hope that helps?
  
  Posted from TSR Mobile
  13. (Original post by L'Evil Fish)
  Product Rule:
  
  y = (u)(v)
  
  dy/dx = (u')(v) + (u)(v')
  Why product rule?
  
  (Original post by stuart_aitken)
  Hey, yep it's just the chain rule with three steps:
  
  sin
  e^x
  [...]^3
  
  Multiply by the differential of all of those. Don't forget to include the 2 at the start as well!
  
  Hope that helps?
  So it would be 6[sin(e^x)]^2 mutliplied by the differential of what's inside the brackets which is e^xcos(e^x)? Is my differential inside the bracket correct?
  14. (Original post by L'Evil Fish)
  Product Rule:
  
  y = (u)(v)
  
  dy/dx = (u')(v) + (u)(v')
  
  Okay. Ignore me actually. Wow
  What lol? I'm so confused
  
  you're probably way better at maths than i so.... :/
  15. (Original post by SurpriseMe)
  What lol? I'm so confused
  
  you're probably way better at maths than i so.... :/
  I thought we could use it like this:
  
  2Sin^3(e^x)
  
  But then I realised there wasn't an x after Sin^3.
  
  And yes, your differential outside the bracket is right
  16. (Original post by L'Evil Fish)
  I thought we could use it like this:
  
  2Sin^3(e^x)
  
  But then I realised there wasn't an x after Sin^3.
  
  And yes, your differential outside the bracket is right
  Ohhh yeah that would make sense
  
  thank you for checking as well
  17. (Original post by SurpriseMe)
  Ohhh yeah that would make sense
  
  thank you for checking as well
  No problem, my confidence has been knocked a little so I'm wary of answering but I do it anyway
  18. (Original post by L'Evil Fish)
  No problem, my confidence has been knocked a little so I'm wary of answering but I do it anyway
  haha, well as you can see my confidence isn't exactly tip-top otherwise I wouldn't keep asking how to do questions that I'm getting right
  
  well I don't blame you cos if you gave a wrong answer you might have a billion quotes questioning your abilities lol. That's why I stay far away from even answering gcse question
  19. (Original post by SurpriseMe)
  haha, well as you can see my confidence isn't exactly tip-top otherwise I wouldn't keep asking how to do questions that I'm getting right
  
  well I don't blame you cos if you gave a wrong answer you might have a billion quotes questioning your abilities lol. That's why I stay far away from even answering gcse question
  Fair enough
  
  Yeah, lately I've been getting a bit wrong so I haven't answered in a while
  
  What year are you in now?
  20. (Original post by L'Evil Fish)
  Fair enough
  
  Yeah, lately I've been getting a bit wrong so I haven't answered in a while
  
  What year are you in now?
  
  I'm in year 13, i'm supposed to be an A* student but i think i lack a bit of self-confidence sometimes and i hate being wrong lol :/
  
  TSR Support Team
  
  We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
  
  This forum is supported by:
  Updated: March 14, 2013
  Today on TSR
  
  ### Congratulations to Harry and Meghan!
  
  But did you bother to watch?
  
  Poll
  Useful resources
  
  ### Maths Forum posting guidelines
  
  Not sure where to post? Read the updated guidelines here
  
  ### How to use LaTex
  
  Writing equations the easy way
  
  ### Study habits of A* students
  
  Top tips from students who have already aced their exams
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a series of math questions and answers related to differentiation. The user initially asks for verification of their solution to a differentiation problem and then proceeds to ask for help with another differentiation problem. The community members provide feedback and confirm the correctness of the user's solutions.
  
  Label: multiple-question-and-answer