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Implicit differentiation

  • 19-04-2012 09:42PM
    #1
    Inbox
    Closed Accounts Posts: 810 ✭✭✭


    Hi I'm tackling some new problems. Can somebody please advise me.

    Q. Check that the given values for x and y satisfy the given equation, then use implicit differentiation to calculate dy/dx for these values.

    x=-2 and y=1 for the equation 2y^2-1/y+3x+5 =0

    So I'm wondering do I put the x and y value in at the end? and also what do I do to the 1/y term, what does that change into? Thanks :)


Comments

  • #2
    sullanefc
    Closed Accounts Posts: 57 ✭✭sullanefc


    Inbox wrote: »
    Hi I'm tackling some new problems. Can somebody please advise me.

    Q. Check that the given values for x and y satisfy the given equation, then use implicit differentiation to calculate dy/dx for these values.

    x=-2 and y=1 for the equation 2y^2-1/y+3x+5 =0

    So I'm wondering do I put the x and y value in at the end? and also what do I do to the 1/y term, what does that change into? Thanks :)

    The question is asking you to put in the x and y values in to the equation above to check that it is on the curve first.

    Then differentiate, and put in the x and y values again. (This is usually used to find the slope of the tangent to the curve at that point).

    On the 1/y term, write it as y^-1 and then differentiate as you normally would for implicit differentiation.


  • #3
    Inbox
    Closed Accounts Posts: 810 ✭✭✭Inbox


    Thank you :D


  • #4
    joeperry
    Closed Accounts Posts: 527 ✭✭✭joeperry


    Hi i have one here that i'm finding tricky.

    Use implicit differentiation too calculate dy/dx

    2xy-ln(3y)=2x^3+y^2

    The answer i got is,

    6x^2-2y / 2x-1/3y-2y


    The ln(3y) is what im unsure of, i'm thinking it becomes 1/3y? Thanks


  • #5
    TheBody
    Registered Users, Registered Users 2 Posts: 5,657 ✭✭✭TheBody


    joeperry wrote: »
    Hi i have one here that i'm finding tricky.

    Use implicit differentiation too calculate dy/dx

    2xy-ln(3y)=2x^3+y^2

    The answer i got is,

    6x^2-2y / 2x-1/3y-2y


    The ln(3y) is what im unsure of, i'm thinking it becomes 1/3y? Thanks

    Hi Joe,

    Yea, you have a small error in the derivative of the [latex]\ln(3y)[/latex] bit.

    If you think about this:
    [latex]\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}[/latex].

    So in your case:
    [latex]\frac{d}{dx}\ln(3y)=\frac{3}{3y}\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}[/latex]

    Hope that makes sense!!


  • #6
    joeperry
    Closed Accounts Posts: 527 ✭✭✭joeperry


    Thanks, i still not totally sure but i'll battle on. Hey how are you typing out the maths bit. What program is that? Thanks.


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  • #7
    joeperry
    Closed Accounts Posts: 527 ✭✭✭joeperry


    Sorry just a quick one if i had ln(2-3x) does that become 1/2-3x? Wolframalpha is telling me it's 3/3x-2. :confused:


  • #8
    TheBody
    Registered Users, Registered Users 2 Posts: 5,657 ✭✭✭TheBody


    joeperry wrote: »
    Sorry just a quick one if i had ln(2-3x) does that become 1/2-3x? Wolframalpha is telling me it's 3/3x-2. :confused:

    [latex]\frac{d}{dx}\ln(2-3x)=\frac{-3}{2-3x}[/latex] (which is the same thing as Wolframalpha with a little manipulation).

    The derivative of the log of a function is the derivative of the function over the original function. It's easy enough to prove using the chain rule.


  • #9
    TheBody
    Registered Users, Registered Users 2 Posts: 5,657 ✭✭✭TheBody


    joeperry wrote: »
    Thanks, i still not totally sure but i'll battle on. Hey how are you typing out the maths bit. What program is that? Thanks.

    I'm typing the maths stuff using a typesetting code called Latex. There is a whole forum on it here:

    http://www.boards.ie/vbulletin/forumdisplay.php?f=1286

    It is amazingly powerful for writing maths stuff but it does take some effort to learn it.


  • #10
    joeperry
    Closed Accounts Posts: 527 ✭✭✭joeperry


    OK thanks i'm getting it now ;)


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