Results 1 to 3 of 3

Math Help - Help with a few basic derivatives...

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    9

    Help with a few basic derivatives...

    1. y = (x^2 + 1)^3 * (x^3 - 1)^2

    2. y = (x^3 + 1) / (x^3 - 2)

    3. y = x^2 * sqrt(x + 1)

    4. find dy/dx and d^2y/dx^2 for x^(1/3) + y^(1/3) = 3 at the point P(1,8)

    If you could briefly explain the methods that would be great too. These are the only ones I am stuck on...thanks1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by golfman44 View Post
    1. y = (x^2 + 1)^3 * (x^3 - 1)^2

    2. y = (x^3 + 1) / (x^3 - 2)

    3. y = x^2 * sqrt(x + 1)

    4. find dy/dx and d^2y/dx^2 for x^(1/3) + y^(1/3) = 3 at the point P(1,8)

    If you could briefly explain the methods that would be great too. These are the only ones I am stuck on...thanks1

    (uv)'=u'v+uv'\,\,\,and\,\,\,\left(\frac{u}{v}\righ  t)'=\frac{u'v-uv'}{v^2} for any two differentiable functions, so for example

    \frac{x^3+1}{x^3-2}=1+\frac{3}{x^3-2}\,\,\,so\,\,\left(1+\frac{3}{x^3-2}\right)'=\frac{-9x^2}{\left(x^3-2\right)^2}

    About the last one: differentiate implicitly every summand wrt x and y:

    \frac{1}{3}x^{-\frac{2}{3}}dx+ \frac{1}{3}y^{-\frac{2}{3}}dy=0\Longrightarrow \frac{dy}{dx}=\left(\frac{y}{x}\right)^{\frac{2}{3  }} and now plug in (1,8)

    For the second derivative \frac{dy^2}{dx^2} do as above with every summand.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    40
    Quote Originally Posted by tonio View Post

    About the last one: differentiate implicitly every summand wrt x and y:

    \frac{1}{3}x^{-\frac{2}{3}}dx+ \frac{1}{3}y^{-\frac{2}{3}}dy=0\Longrightarrow \frac{dy}{dx}=\left(\frac{y}{x}\right)^{\frac{2}{3  }} and now plug in (1,8)

    For the second derivative \frac{dy^2}{dx^2} do as above with every summand.

    Tonio
    I think Tonio made a small mistake: the derivative for this last one should be \frac{dy}{dx} = - (\frac{y}{x})^{\frac{2}{3}}

    Because this is actually a function, there is no need to differentiate implicitly, as the equation can be rearranged to find y explicitly in terms of y:

    y = (3 - x^{\frac{1}{3}})^3 = 27 - 27x^{\frac{1}{3}} + 9x^{\frac{2}{3}} - x

    Then: \frac{dy}{dx} = -9x^{\frac{-2}{3}} + 6x^{\frac{-1}{3}} - 1. Now just sub in x = 1.

    For \frac{d^2 y}{dx^2}: \frac{d^2 y}{dx^2} = 6x^{\frac{-5}{3}} - 2x^{\frac{-4}{3}} . Again sub in x=1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Having trouble with basic Derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 8th 2010, 05:33 PM
  2. Very basic derivatives: 1/(1-x) and x/(1-x)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 9th 2010, 08:10 AM
  3. Basic question - derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 17th 2010, 03:55 PM
  4. where are my mistakes? (basic derivatives)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 23rd 2009, 06:26 AM
  5. basic derivatives problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 5th 2007, 10:34 AM

Search Tags


/mathhelpforum @mathhelpforum