Hey there! Sign in to join this conversationNew here? Join for free

Implicit differentiation, what am I doing wrong? Watch

    • Thread Starter
    Offline

    1
    ReputationRep:
    Use implicit differentiation to find dy/dx for cos(x+y) + sin(x+y) = 0.5

    Differentiating this I got -sin(x+y)(1+dy/dx) + cos(x+y)(1+dy/dx) = 0

    I rearranged this to get dy/dx = sin(x+y) - cos(x+y) / cos(x+y) - sin (x+y)

    Then using the given equation I rearranged and got

    -cos (x+y) = -0.5 +sin(x+y)
    and -sin(x+y) = -0.5 + (cosx+y)
    and tried subbing in my dy/dx but it gives some other expression, the answer should just be -1

    where do I go from here?

    Thanks
    • Study Helper
    Offline

    13
    (Original post by DonWorryJockIsHere)
    Use implicit differentiation to find dy/dx for cos(x+y) + sin(x+y) = 0.5

    Differentiating this I got -sin(x+y)(1+dy/dx) + cos(x+y)(1+dy/dx) = 0
    Try factorising at this point.

    Edit: You don't actually need implicit differentiation for this question, and could use the sum/product trig. rules, if you've covered them.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by ghostwalker)
    Try factorising at this point.

    Edit: You don't actually need implicit differentiation for this question, and could use the sum/product trig. rules, if you've covered them.
    I factorised dy/dx and made it the subject as thats what you are trying to find?
    • Community Assistant
    Offline

    19
    ReputationRep:
    (Original post by DonWorryJockIsHere)
    I factorised dy/dx and made it the subject as thats what you are trying to find?
    Factorising where ghostwalker indicated gives you the solution you seek immediately.
    Offline

    21
    ReputationRep:
    (Original post by DonWorryJockIsHere)
    Use implicit differentiation to find dy/dx for cos(x+y) + sin(x+y) = 0.5

    Differentiating this I got -sin(x+y)(1+dy/dx) + cos(x+y)(1+dy/dx) = 0

    I rearranged this to get dy/dx = sin(x+y) - cos(x+y) / cos(x+y) - sin (x+y)

    Then using the given equation I rearranged and got

    -cos (x+y) = -0.5 +sin(x+y)
    and -sin(x+y) = -0.5 + (cosx+y)
    and tried subbing in my dy/dx but it gives some other expression, the answer should just be -1

    where do I go from here?

    Thanks
    The answer is a lot easier than you are trying to make it:
    1.first differentiate in respect to x remembering that cos(ax+b)=-a sin(ax+b) and sin(ax+b)=a cos(ax+b) we get cos(x+y)-sin(x+y)
    differentiate with respect to y because there is only one x and y, we will get the same result cos(x+y)-sin(x+y)
    3.the next step is to add a minus sign and divide you answers to parts 1 and 2, which are the same giving you -1.

    This method is a much simpler method of implicit differentiation that they don't seem to give in the textbooks for some reason.
    • Community Assistant
    Offline

    19
    ReputationRep:
    Dalek isn't trolling in case people are suspicious. He did fail to say that you need to make the implicit equation equal to zero before using this technique.

    I can't agree that this is any easier than ghostwalker's method though.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Mr M)
    Dalek isn't trolling in case people are suspicious. He did fail to say that you need to make the implicit equation equal to zero before using this technique.

    I can't agree that this is any easier than ghostwalker's method though.
    Is it just (1+dy/dx) (cos(x+y) - sin (x+y)) = 0? then you can get dy/dx = -1? but I dont see why cos (x+y) - sin (x+y) = 0?
    • Study Helper
    Offline

    13
    (Original post by DonWorryJockIsHere)
    Is it just (1+dy/dx) (cos(x+y) - sin (x+y)) = 0? then you can get dy/dx = -1? but I dont see why cos (x+y) - sin (x+y) = 0?
    If ab=0, then either a=0 or b= 0 or both.

    In this case

    either dy/dx=-1,

    or cos(x+y)=sin(x+y) in which case tan(x+y)=1, i.e is constant. So x+y is constant. I.e. x+y=c and y=c-x, and once again dy/dx = -1.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.