Results 1 to 6 of 6

Math Help - Differentiation structured questions

  1. #1
    Junior Member
    Joined
    Nov 2010
    Posts
    50

    Differentiation structured questions

    Hey,

    I need help with some questions that I have been stuck with on my crash course to teach myself calculus before school starts.

    1) If  y = x +\sqrt{x^2 -9} , show that

     \displaystyle (x^2 -9) \dfrac{d^2y}{dx^2} + x \dfrac{dy}{dx} - y = 0

    I have tried to solve for the derivatives of y and I derived extremely complicated fractions with powers of 3/2 and such and I cannot simplify the equation to = 0.


    2)
    a)  \dfrac {d}{dx} \dfrac{3x-4}{x+7} = \dfrac {25}{(x+7)^2} (found)

    b) Using your answer in (a) , differentiate the following with respect to x
     \displaystyle y = sin(\dfrac{3x-4}{x+7})^8

    For this, I am unsure of what to let be u so that I can use the chain rule in the form
     \dfrac{du}{du}  \, \dfrac{du}{dx} or  \dfrac{dsinu}{du}  \, \dfrac{du}{dx}
    I have dealt with trigo functions like sin^8(x) and sin(8x) but never in the form that the question gave (8th power only applies to the fraction and not the whole function) and I do not know how to proceed.

    Any help at all is appreciated
    Last edited by arccos; December 14th 2010 at 12:58 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, what did you get for the derivatives in 1)?

    For 2) b), you have to use the chain rule from the outside in, and you're going to have to use it twice (having done part a means you don't have to do it three times, like you normally would). Here's the first step:

    \displaystyle \frac{dy}{dx} = \cos\left(\left(\dfrac{3x-4}{x+7}\right)^{\!\!8}\right)\;\frac{d}{dx}\left(\  left(\dfrac{3x-4}{x+7}\right)^{\!\!8}\right).

    Thank you for the sentence clarifying what the exponent 8 applies to, but if you include extra parentheses like I've done here, you don't need to bother.

    Can you keep going from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2010
    Posts
    50
    Quote Originally Posted by Ackbeet View Post
    Well, what did you get for the derivatives in 1)?

    For 2) b), you have to use the chain rule from the outside in, and you're going to have to use it twice (having done part a means you don't have to do it three times, like you normally would). Here's the first step:

    \displaystyle \frac{dy}{dx} = \cos\left(\left(\dfrac{3x-4}{x+7}\right)^{\!\!8}\right)\;\frac{d}{dx}\left(\  left(\dfrac{3x-4}{x+7}\right)^{\!\!8}\right).

    Thank you for the sentence clarifying what the exponent 8 applies to, but if you include extra parentheses like I've done here, you don't need to bother.

    Can you keep going from here?
    Thanks for the tip. 2 is solved.
    For qn 1 ,
     \dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{x^2 -9}}

     \dfrac{d^2y}{dx^2} = \dfrac{1}{\sqrt{x^2-9}} - \dfrac{x^2}{(x^2-9)^{\frac{3}{2}}}

    All the weird exponents make simplifying hard. Have i gone wrong with the deriviation?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,368
    Thanks
    1313
    A^{3/2}= A(A^{1/2})= A\sqrt(A) so that could be written
    \frac{d^2y}{dx^2}= \frac{1}{\sqrt{x^2- 9}}- \frac{x^2}{(x^2-9)\sqrt{x^2- 9}}
    and then get "common denominators by multiplying the numerator and denominator of the first fraction by x^2- 9:
    \frac{d^2y}{dx^2}= \frac{x^2- 9}{(x^2-9)\sqrt{x^2- 9}}- \frac{x^2}{(x^2- 9)\sqrt{x^2- 9}}
    \frac{d^2y}{dx^2}= -\frac{9}{(x^2- 9)\sqrt{x^2- 9}}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2010
    Posts
    50
     \dfrac{-9}{\sqrt{x^2-9}} + \dfrac{x^2}{\sqrt{x^2-9}} ....

    In the end, the 9 and x^2 canceled out,I crossed out sqrt x^2-9 from both the denominator and numerator but im left with -x...

    Sorry about the shortcut,i screwed up my latex.

    EDIT:
    Oh I have solved the problem. The thing I overlooked was to not simplify the numerator but to group them together.
    \dfrac{-9}{\sqrt{x^2-9}} + \dfrac{x^2}{\sqrt{x^2-9}} 's numerator can be regrouped to x^2 - 9 which cancels out with the other x^2 - 9 , leaving 0 which divided by anything,is still = 0 = RHS.

    Thanks for the help anyway guys. I appreciate it
    Last edited by arccos; December 14th 2010 at 07:52 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Differentiating and rearranging gives:

    \displaystyle {y = x +\sqrt{x^2 -9}}} \Rightarrow y' = 1+\frac{x}{\sqrt{x^2-9}} \Rightarrow  y' -\frac{x}{\sqrt{x^2-9}}= 1

    But x = y-\sqrt{x^2-9}, so we have:

    \displaystyle \Rightarrow y'-\frac{y-\sqrt{x^2-9}}{\sqrt{x^2-9}} = 1 \Rightarrow y'-\frac{y}{\sqrt{x^2-9}} = 0 ~ ~ \;\; (\star)

    Differentiate this, noting of course that:

    \displaystyle \left(\frac{y}{\sqrt{x^2-9}}\right)' = \frac{y'\sqrt{x^2-9}-y\left(\sqrt{x^2-9}\right)'}{(x^2-9)} = \frac{y'\left(x^2-9\right)-yx}{(x^2-9)\sqrt{x^2-9}}

    So we have:

    \displaystyle y''-\frac{y'\left(x^2-9\right)-yx}{(x^2-9)\sqrt{x^2-9}} = 0 \Rightarrow y''-\frac{y'(x^2-9)}{(x^2-9)\sqrt{x^2-9}}-x\frac{y}{(x^2-9)\sqrt{x^2-9}} = 0<br />

    \displaystyle \Rightarrow (x^2-9)y''-\frac{y'(x^2-9)}{\sqrt{x^2-9}}-x\frac{y}{\sqrt{x^2-9}} = 0 \Rightarrow (x^2-9)y''-y'\sqrt{x^2-9}-x\frac{y}{\sqrt{x^2-9}} = 0

    But from \left(\star\right) we have y'\sqrt{x^2-9} = y and \frac{y}{\sqrt{x^2-9}} = -y', so indeed \displaystyle \boxed{(x^2-9)y''+xy'-y = 0}.
    Last edited by TheCoffeeMachine; December 15th 2010 at 05:27 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differentiation Questions
    Posted in the Calculus Forum
    Replies: 13
    Last Post: February 13th 2011, 12:38 PM
  2. Differentiation questions.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 6th 2010, 05:48 AM
  3. Differentiation questions.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 9th 2010, 08:46 PM
  4. Differentiation Questions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 26th 2010, 08:32 AM
  5. differentiation questions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 27th 2009, 03:31 PM

Search Tags


/mathhelpforum @mathhelpforum