Results 1 to 4 of 4

Math Help - derivatives

  1. #1
    Newbie
    Joined
    Jun 2007
    Posts
    2

    derivatives

    Is anyone up for helping me out with a bit of homework?
    Pretty please???

    I've got 3 differentiation questions and 3 simultaneous equations, and I just can't get my head round them. I've tried for 2 weeks now, and I just can't do it. I feel sick to the stomach for having to ask, but I'll mess up a whole years course if I can't do them. Here goes.................

    Find dy/dx for:
    A. y = x(cubed) - x(squared)

    B. y = (x-1) (x+2)

    Find the turning points for the following and state what type of turning points they are:
    C. y = x(cubed) - 3x + 2

    Solve for x and y:
    D. 17x + 9y = 20, 5x - 2y = -22

    E. x = 2y, x(squared) + 3xy = 10

    F. y - x = 2, 2x(squared) + 3xy = y(squared) = 8

    I'm doing a psychology course (believe it or not), so if anyone needs any help with that sort of stuff, I'm more than happy to help.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by buddhabum View Post
    Is anyone up for helping me out with a bit of homework?
    Pretty please???

    I've got 3 differentiation questions and 3 simultaneous equations, and I just can't get my head round them. I've tried for 2 weeks now, and I just can't do it. I feel sick to the stomach for having to ask, but I'll mess up a whole years course if I can't do them. Here goes.................
    calm down, take deep breaths. now think happy thoughts. are you calm now? ok, let's get started.

    Find dy/dx for:
    A. y = x(cubed) - x(squared)
    For this question, we need what's called the Power Rule for derivatives. The rule says:

    \frac {d}{dx} x^n = n x^{n - 1}

    That is, to take the derivative of a variable raised to some power, we multiply by the power and then subtract one from the power to get the new power. now let's see how this applies to your question

    y = x^3 - x^2

    \Rightarrow \frac {dy}{dx} = 3x^{3 - 1} - 2x^{2 - 1} .......we usually skip this step, but i just wanted to show you the rule in action

    \Rightarrow \frac {dy}{dx} = 3x^2 - 2x

    B. y = (x-1) (x+2)
    this is more complicated. now you can expand everything and use the power rule, but if you want to look sophisticated you would use what is called the Product Rule on this one.

    The Product Rule says:

    If we have the product of two functions, say f(x) and g(x), and we want to take the derivative of this product, it will be the derivative of the first function times the second added to the derivative of the second function times the first. That is,

    \frac {d}{dx} \left( f(x) \cdot g(x) \right) = f'(x) \cdot g(x) + f(x) \cdot g'(x)

    Now let's see this rule in action.

    y = (x - 1)(x + 2)

    \Rightarrow \frac {dy}{dx} = 1 \cdot (x + 2) + (x - 1) \cdot 1 ........now simplify

    \Rightarrow \frac {dy}{dx} = 2x + 1

    Find the turning points for the following and state what type of turning points they are:
    C. y = x(cubed) - 3x + 2
    i'll let you practice with this one. I will give you this hint. Turning points occur when \frac {dy}{dx} = 0

    afterwards, we find \frac {d^2 y}{dx^2}, that means the second derivative. you just take the derivative of the \frac {dy}{dx} function.

    Then we plug in the x's we found from setting \frac {dy}{dx} = 0 into this new function. and we can tell what the turning point is by the following:

    The turning point is a local maximum if we get \frac {d^2 y}{dx^2} < 0 when we plug in a critical value (that is the value that makes dy/dx = 0)

    The turning point is a local minimum if we get \frac {d^2 y }{dx^2} > 0 when we plug in a critical value

    The turning point is POSSIBLY an inflection point if we get \frac {d^2 y}{dx^2} = 0 when we plug in a critical value. Further tests are usually needed in this case.

    I'm doing a psychology course (believe it or not), so if anyone needs any help with that sort of stuff, I'm more than happy to help.
    I've never seen anyone ask a Psychology question here, so you might not get to help anybody, just letting you know


    if you have any questions, don't hesitate to ask
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2007
    From
    el paso, TX
    Posts
    30
    wow... D, E and F dont seem to match the first three :P, anyway to solve

    17x +9y = 20
     5x -2y = -22

    multiply the top equation by 2, and the bottom by 9
     34x + 18y = 40
     45 x -18y = -22(9)
    now add the two,
     79x = 40  -22(9), voila that easy enough..

    for E lets use substitution instead, since x = 2y then we can replace all the x's in the second equation... to get (2y)^2 + 6y^2 = 10... i leave the rest to you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2007
    Posts
    2
    Oh WOW!!! What a pair of sweeties! Still a tad confused, but I'm sure a certain can of energy drink will help. Right.......I'll print off all the info you darlings have given me and I'll get back to you in the morning (its just gone midnight here).

    Thank you so, so, so, so, much! The sinking feeling is melting away and I really can't say thank you enough!!!!

    You really are angels, I've gone from to in half an hour.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives and Anti-Derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 6th 2011, 06:21 AM
  2. Replies: 1
    Last Post: July 19th 2010, 04:09 PM
  3. Derivatives with both a and y
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2009, 09:17 AM
  4. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  5. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum