1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding derivative and solve for t

  1. Oct 27, 2011 #1
    Hello so I'm supposed to find the derivative of (-t2+4)/(t2+4)2

    and using the quotient rule + the chain rule I got this mess:
    dy/dx= -2t(t^2+4)^2-2t(2t2+8)(t2+4)2/ (t2+4)4

    and now I have to set that whole thing to 0. I know the numerator must be equal to 0 which mean that one of the solution is t=0. Is there any other solutions?Can you help me simplify/factor it to find the other solution?Thank you
  2. jcsd
  3. Oct 27, 2011 #2


    User Avatar
    Science Advisor
    Homework Helper

    Your 'mess' is a ways from being right yet. Once you get it right I'd start with canceling out the common factors of (t^2+4) in the numerator and denominator. It will look a lot less messy.
  4. Oct 28, 2011 #3


    Staff: Mentor

    Since there is no x in your original expression, your derivative shouldn't be dy/dx. Also, there's no y in that expression, either, so where did it come from?

    Why do you think you need to set the derivative to 0? The question merely asks you to find the derivative.
  5. Oct 28, 2011 #4
    Hello there,
    the function is v(t)=-t2+4/(t2+4)2
    and to find the derivative I'm suppose to find v'(t) or a(t)
    The reason why I have to set the second derivative to 0 so that I could find the point at which the function is concave up or down. (those are called stationary points?)
    I find the second derivative by using the equation:
    and there is a comment said that my second derivative is wrong. Can you guide me where I make the mistake?
    Thank you
  6. Oct 28, 2011 #5
    It could have gone wrong anywhere. Show us your steps in simplifying your derivative after applying the quotient rule. Only then will we be able to pinpoint your mistake. So when you set your 2nd derivative = 0, you are looking at when there is constant velocity or 0 acceleration. I wouldn't call these stationary points since the object is still moving in some direction. Stationary points would only be when v(t)=0
  7. Oct 28, 2011 #6
    oh okay here is the step
    a(t)= -2t(t^2+4)^2 is the part f'g and then subtract the g'f part I used the chain rule to find g' which is 2t(2t^2+8)<----g'(t^2+4)^2<-----f and divide by g^2 which is( (t^2+4)^2)^2) which clean up to be equal (t^2+4)^4

    and that why I got -2t(t^2+4)^2-2t(2t^2+8)(t^2+4)^2/ (t^2+4)^4
  8. Oct 28, 2011 #7


    User Avatar
    Science Advisor
    Homework Helper

    The f'g part looks ok. The g'f part doesn't. It looks more like g'g to me.
  9. Oct 28, 2011 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    (You really do need to learn to use parentheses where they're needed.)

    I take it that you have f(t)/g(t) = (-t2+4)/(t2+4)2 .

    Then [itex]\displaystyle \frac{d}{dt}\left(\frac{f(t)}{g(t)}\right)=\frac{f\,'\!(t)\,g(t)-f(t)\,g'(t)}{(g(t))^2}[/itex]

    I don't see f(t) in the second term of the numerator in your expression.
  10. Oct 29, 2011 #9


    Staff: Mentor

    This information is part of the problem statement, which should have been in your first post.
  11. Oct 29, 2011 #10
    Hello everyone, so I redo the derivative

    a(t)= -2t(t2+4)2-2(t2+4)(2t)(-t2+4)/ (t2+4)4

    does that look about right?
  12. Oct 29, 2011 #11


    Staff: Mentor

    Like Sammy said, you need more parentheses.

    What you wrote would be interpreted as
    [tex]a(t) = -2t(t^2 + 4)^2 - \frac{2(t^2 + 4)(2t)(-t^2 + 4)}{(t^2 + 4)^4}[/tex]

    You need another pair of parentheses around the entire numerator.

    After making that fix, that looks to be correct, but if the answer is given, it probably won't look like this. In the textbook answers, they usually pull out common factors in the numerator to write the numerator as a product of factors rather than as a sum or difference of terms.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook