Results 1 to 4 of 4

Math Help - Implicit Differentiation

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    7

    Implicit Differentiation

    I think my teacher was talking about turning y into y(x) or something but I'm just completely lost on this one
    Attached Thumbnails Attached Thumbnails Implicit Differentiation-5.bmp  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    5x^2+4x+xy=2 using implicit we get

    10x+4+y+xy'=0

    y(2)=-13

    Solving for y' we get

    y'=- \left( \frac{10x+4+y}{x} \right)

    evaluate at x=2

    y'(2)=-\left( \frac{10(2)+4+(-13)}{2} \right)=-\frac{11}{2}=5.5
    Last edited by TheEmptySet; March 25th 2008 at 10:43 PM. Reason: wrong value of x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Let's do an example to illustrate implicit differentiation:

    x^{2} + y^{3} = 9

    Now, if you were to ask to find the derivative of this, you could solve for y then differentiate the equation with respect to x. However, not all functions are easily represented only in terms of x (ex. you could not solve for y for this equation: x^{2} + xy + y^{2} = 5). Fortunately, we are still able to find the derivative to such functions.

    Note that y is a function of x so that we can differentiate it normally. However, we would have to use the chain rule. I think if you study the example, you'll get what I mean:

    x^{2} + y^{3} = 9
    2x + \underbrace{3y^{2}y'}_{} = 0 \quad \mbox{Differentiated with respect to x}

    Looking at the underbrace, notice how we used the power rule (3y^2) and in addition the chain rule (hence the y'). Continuing on to solve for y':
    3y^{2}y' = -2x
    y' = \frac{-2x}{3y^{2}}

    Note it's perfectly fine to have both x and y in our solution to y'.

    That is basically the trick when implicitly differentiating equations. See if you can extend this to your question.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by lyzzi3 View Post
    I think my teacher was talking about turning y into y(x) or something but I'm just completely lost on this one
    5x^2+4x+xy(x)=2
    10x+4+y(x)+x[y'(x)]=0

    If, y(2)=-13

    10(2)+4+(-13)+(2)[y'(2)]=0
    11+2[y'(2)]=0
    y'(2)=-\frac{11}{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 13th 2010, 09:58 AM
  3. implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 1st 2010, 07:42 AM
  4. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 28th 2010, 03:19 PM
  5. Implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 26th 2010, 05:41 PM

Search Tags


/mathhelpforum @mathhelpforum