Results 1 to 5 of 5

Math Help - Using implicit differentiation to find the slope of the tangent line

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    3

    Using implicit differentiation to find the slope of the tangent line

    Stuck on this problem..



    If it were using explicit differentiation, I could do this, but implicitly I'm having some trouble.

    What I tried was implicitly differentiating both sides:

    (d/dx) [ (4x + 3y)^(1/2) + (2xy)^(1/2) ] = 0

    and after differentiating, I ended up with with:

    (1/2)*(4x + 3y) * (4+3y') + 2(y + xy') = 0

    From there, I plugged in the x,y values (3,8) for x and y, then solved for y', getting..

    18 * (4 + 3y') + 2(8 + 3y') = 0
    60y' = -88
    y = -88/60

    which is incorrect.

    Am I approaching the problem from the wrong angle?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Using implicit differentiation to find the slope of the tangent line

    Your square roots magically disappeared.

    \frac{d}{dx}\sqrt{x}\;=\;\frac{1}{2\cdot\sqrt{x}}

    Still a square root in there!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2011
    Posts
    3

    Re: Using implicit differentiation to find the slope of the tangent line

    Aha, thank you much!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2011
    Posts
    3

    Re: Using implicit differentiation to find the slope of the tangent line

    Nevermind, please delete.
    Last edited by Chronoc2005; September 11th 2011 at 09:11 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,513
    Thanks
    1404

    Re: Using implicit differentiation to find the slope of the tangent line

    Quote Originally Posted by Chronoc2005 View Post
    Stuck on this problem..



    If it were using explicit differentiation, I could do this, but implicitly I'm having some trouble.

    What I tried was implicitly differentiating both sides:

    (d/dx) [ (4x + 3y)^(1/2) + (2xy)^(1/2) ] = 0

    and after differentiating, I ended up with with:

    (1/2)*(4x + 3y) * (4+3y') + 2(y + xy') = 0

    From there, I plugged in the x,y values (3,8) for x and y, then solved for y', getting..

    18 * (4 + 3y') + 2(8 + 3y') = 0
    60y' = -88
    y = -88/60

    which is incorrect.

    Am I approaching the problem from the wrong angle?
    \displaystyle \begin{align*}\sqrt{4x + 3y} + \sqrt{2xy} &= 12.9 \\ \left(4x + 3y\right)^{\frac{1}{2}} + \left(2xy\right)^{\frac{1}{2}} &= 12.9 \\ \frac{d}{dx}\left[\left(4x + 3y\right)^{\frac{1}{2}} + \left(2xy\right)^{\frac{1}{2}}\right] &= \frac{d}{dx}\left(12.9\right) \\ \frac{d}{dx}\left[\left(4x + 3y\right)^{\frac{1}{2}}\right] + \frac{d}{dx}\left[\left(2xy\right)^{\frac{1}{2}}\right] &= 0 \end{align*}

    The first of these derivatives...

    \displaystyle z = \left(4x + 3y\right)^{\frac{1}{2}}

    Let \displaystyle u = 4x + 3y \implies z = u^{\frac{1}{2}}.

    \displaystyle \frac{du}{dx} = 4 + 3\frac{dy}{dx}.

    \displaystyle \begin{align*} \frac{dz}{du} &= \frac{1}{2}u^{-\frac{1}{2}} \\ &= \frac{1}{2\sqrt{4x + 3y}} \end{align*}

    So \displaystyle \frac{dz}{dx} = \frac{4 + 3\frac{dy}{dx}}{2\sqrt{4x + 3y}}


    The second of these derivatives.

    \displaystyle w = \left(2xy\right)^{\frac{1}{2}}

    Let \displaystyle u = 2xy \implies w = u^{\frac{1}{2}}.

    \displaystyle \frac{du}{dx} = 2y + 2x\,\frac{dy}{dx}.

    \displaystyle \begin{align*} \frac{dw}{du} &= \frac{1}{2}u^{-\frac{1}{2}} \\ &= \frac{1}{2\sqrt{2xy}} \end{align*}

    So \displaystyle \frac{dw}{dx} = \frac{2y + 2x\,\frac{dy}{dx}}{2\sqrt{2xy}}


    Therefore, going back to your original equation...

    \displaystyle \begin{align*} \frac{d}{dx}\left[\left(4x + 3y\right)^{\frac{1}{2}}\right] + \frac{d}{dx}\left[\left(2xy\right)^{\frac{1}{2}}\right] &= 0 \\ \frac{4 + 3\frac{dy}{dx}}{2\sqrt{4x + 3y}} + \frac{2y + 2x\,\frac{dy}{dx}}{2\sqrt{2xy}} &= 0 \\ \frac{\left(4 + 3\frac{dy}{dx}\right)\sqrt{2xy} + \left(2y + 2x\,\frac{dy}{dx}\right)\sqrt{4x + 3y}}{2\sqrt{4x + 3y}\sqrt{2xy}} &= 0 \\ \left(4 + 3\frac{dy}{dx}\right)\sqrt{2xy} + \left(2y + 2x\,\frac{dy}{dx}\right)\sqrt{4x + 3y} &= 0 \\ 4\sqrt{2xy} + 3\sqrt{2xy}\,\frac{dy}{dx} + 2y\sqrt{4x + 3y} + 2x\sqrt{4x + 3y}\,\frac{dy}{dx} &= 0 \\ \left(3\sqrt{2xy} + 2x\sqrt{4x + 3y}\right)\frac{dy}{dx} &= -4\sqrt{2xy} - 2y\sqrt{4x + 3y} \\ \frac{dy}{dx} &= \frac{-2\left(2\sqrt{2xy} + y\sqrt{4x + 3y}\right)}{3\sqrt{2xy} + 2x\sqrt{4x + 3y}}\end{align*}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 28th 2012, 03:27 PM
  2. tangent line, implicit differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 19th 2010, 09:50 PM
  3. implicit differentiation, tangent line help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 17th 2010, 02:48 PM
  4. implicit differentiation to tangent line
    Posted in the Calculus Forum
    Replies: 15
    Last Post: November 5th 2009, 10:32 AM
  5. Implicit Differentiation Slope Tangent Line
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 19th 2008, 08:28 PM

Search Tags


/mathhelpforum @mathhelpforum