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Derivative using quotient and product rules

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data
    y=[itex]\frac{}{}tsint/1+t[/itex]


    2. Relevant equations



    3. The attempt at a solution
    What I do (and arrive at the wrong solution) is the fallowing. Can you please tell me where i go wrong. *I am going to leave out the denominator as i go through the problem because it stays the same the whole time and will make the problem look messier.

    First step: I used the quotient rule and product rule for derivatives for the first step. Quotent rule because the problem is a fraction and quotent rule because the top is a product. after taking the necessary derivatives I get,
    y'=1+t[tcos(t)+sin(t)]-tsin(t)
    y'=tcos(t)+sin(t)+t^2cos(t)+tsin(t)-tsin(t)
    y'=tcos(t)+sin(t)+t^2cos(t)/(1+t)^2
    That answer is incorrect. the correct answer is y'=(t^2+t)cos(t)+sin(t)/(1+t)^2

    Help appreciated!
     
  2. jcsd
  3. Mar 17, 2013 #2
    Is that y= tsin(t)/(1+t) or y=(tsin(t)/1) + t?

    I imagine the former but your comment about the denominator staying the same makes me believe you think it is the latter?
     
  4. Mar 17, 2013 #3

    Ray Vickson

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    I cannot parse your function, due to your lack of parentheses. Do you mean
    [tex] y = t \sin(t) + t \text{ (which is what you wrote), }\\
    \text{or}\\
    y = t \sin\left( \frac{t}{t+1}\right),\\
    \text{or}\\
    y = \frac{t \sin t}{t+1}?[/tex]
     
  5. Mar 17, 2013 #4

    SteamKing

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    If you take the derivative of t*sin(t), how in the world to you get t^2 in the result?

    Since this is a product, write down what is u, what is v, and what u' and v' are.
    Then combine according to the chain rule. Trying to do this on the fly has left you confused.
     
  6. Mar 17, 2013 #5
    yes, sorry the equation is tsin(t)/(1+t)

    SteamKing: We have not learned the chain rule yet, it is the topic of Mondays lecture. But i didn't get t^2 as the derivative of tsin(t). When I take the derivative of tsin(t) I get
    tcos(t)-tsin(t). I incorrectly factored out a t for some reason. But I don't think that will fix my problem. It was just a typo, sorry.
     
  7. Mar 17, 2013 #6

    vela

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    After deciphering what you wrote, I'll note you did it correctly, but you really need to quit being so sloppy with your notation. I hope you don't turn work like this in on your homework and exams. If you do, you probably drive your instructor up the wall.

    In your first line, you wrote ##y' = 1+t[t\cos(t)+\sin(t)]-t\sin(t)##. First off, you weren't writing down y'. You were only writing down the numerator, so you shouldn't have said "y'=…". Second, 1+t should have been in parentheses. The numerator of the derivative, correctly written, is ##(1+t)(t \cos t + \sin t) - t\sin t##, which is apparently what you meant based on what you wrote in the second line.

    To get y', all you have to now do is divide by the denominator squared, so
    $$y' = \frac{(1+t)(t \cos t + \sin t) - t\sin t}{(1+t)^2}.$$ If you multiply out the numerator and simplify, you end up with what you got. If you then collect the cosine terms and factor out the cos t, you end up with the other solution. Your answer was right.
     
  8. Mar 20, 2013 #7
    OK thanks.
    This was just my first post placing a math problem on the computer. I'm very neat with my work in class and use brackets and parenthesis when necessary. I also wrote this after many hours of studying so my head was a mess. I will from now on use my dad's computer to write math problems because it has Microsoft word and I can write math problems the way they would look as if they were on paper. I will just copy and paste them on here.
    Good to hear the answer was correct. I guess the program did not take it because it could be simplified more.
     
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