Results 1 to 2 of 2

Math Help - Differentiation Question

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Differentiation Question

    Hi

    I need help on the following questions. I am not sure if the answer i got is correct.

    1)Find the derivative y'(x) of x^4 + y(x^3) + \frac{1}{y(x)^4} - 13 =0

    After going through the process this is what i get:

    \frac{4y^2x^3 + 3y^3x^2 + \frac{dy}{dx}x^3y^2 - 4yx^{-5} - x^{-4} \frac{dy}{dx}}{y^2}

    2)Find the derivative of: \sqrt{\frac{x+8}{(2x+5)(x^2+2}}
    <br />
{\frac{x+8}{(2x+5)(x^2+2}}^{0.5}

    Firstly i used to chain rule and then inside the square root i used the quotient rule.

    Eventually i get this answer:

    {\frac{2x^3-x^2-6x-6}{2\frac{x+8}{(2x+5)(x^2+2)}(2x+5)^2(x^2+2)^2}}

    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, Paymemoney!

    1) Find y'\!:\quad x^4 + x^3y + \frac{1}{x^4y} - 13 \;=\; 0
    Sorry, your answer makes no sense.


    We have: . x^4 + x^3y + x^{-4}y^{-1} - 13 \;=\;0

    Differentiate implicitly: . 4x^3 + x^3(y') + 3x^2y - x^{-4}y^{-2}(y') - 4x^{-5}y^{-1} \;=\;0

    Multiply by x^5y^2\!:\quad 4x^8y^2 + x^8y^2(y') + 3x^7y^3 - x(y') - 4y \;=\;0

    Rearrange terms: . x^8y^2(y') - x(y') \;=\;4y - 4x^8y^2 - 3x^7y^3

    Factor: . x(x^7y^2-1)\cdot y' \;=\;y(4 - 4x^7y - 3x^7y^2)


    Therefore: . y' \;=\;\frac{y(4 - 4x^8y - 3x^7y^2)}{x(x^7y^2-1)}




    2) Diffrerentiate: . y \;=\;\sqrt{\frac{x+8}{(2x+5)(x^2+2)}} \;=\;\left(\frac{x+8}{2x^3+5x^2+4x+10}\right)^{\fr  ac{1}{2}}
    Your game plan is correct, but . . .


    y' \;=\;\frac{1}{2}\left(\frac{x+8}{2x^3+5x^2+4x+10}\  right)^{-\frac{1}{2}}\cdot \left[\frac{(2x^3+5x^2+4x+10)\cdot 1 - (x+8)(6x^2 + 10x + 4)}{(2x^3+5x^2+4x+10)^2}\right]

    y' \;= \;\frac{1}{2}\cdot\frac{(2x^3+5x^2+4x+10)^{\frac{1  }{2}}}{(x+8)^{\frac{1}{2}}}\cdot\left[\frac{-4x^3 - 9x^2 - 80x - 22}{(2x^3+5x^2+4x+10)^2}\right]

    y' \;=\;-\frac{4x^3+9x^2+80x+22}{2(x+8)^{\frac{1}{2}}(2x^3+  5x^2+4x+10)^{\frac{3}{2}}}



    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. ln differentiation question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 7th 2011, 01:47 PM
  2. Replies: 6
    Last Post: July 21st 2010, 06:20 PM
  3. a question about differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 28th 2009, 08:03 AM
  4. Need help with this differentiation question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 7th 2008, 11:29 AM
  5. Differentiation Question
    Posted in the Calculus Forum
    Replies: 20
    Last Post: May 22nd 2008, 07:50 PM

Search Tags


/mathhelpforum @mathhelpforum