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Homework Help: Urgent: One Extremely Basic Derivative Function

  1. Jan 15, 2009 #1


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    1. The problem statement, all variables and given/known data
    Find an equation of the tangent line to the curve y=2x-x^2 at the point (2, -4).

    2. Relevant equations

    3. The attempt at a solution
    I do not have any idea how to do any derivative, other than the equation to do so. Please note that we have been given the answer key - I already know that the solution is y+4=-10(x-2).

    This is for a Precalculus exam (but I've been informed that derivatives are calculus). I have been absent for a while, which is when we learned about derivatives, so any background information would be nice as well, but the test is tomorrow.
  2. jcsd
  3. Jan 15, 2009 #2
    The slop of the tangent can be found by differentiating the given equation, let it be [tex]\frac {\delta y}{\delta x}.[/tex]

    Now, the formula to find the equation is simply,
    [tex]\frac{y-y_1}{x-x_1}=~\frac {\delta y}{\delta x}.[/tex]

    where, [tex] x_1[/tex] and [tex] y_1[/tex] are the given points!

    Substitute, and you'll get the answer, rightaway! :wink:
  4. Jan 15, 2009 #3
    That's only for equations of the form y = m*x + b i.e. linear for some form of linear. In general dy/dx is not that.
  5. Jan 15, 2009 #4


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    Thanks, but could somebody explain basic derivatives in general, please?
  6. Jan 15, 2009 #5
  7. Jan 15, 2009 #6


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    Now I am ;-). I'll post back if I'm still confused. It would be really great if (if anyone has the time) somebody could post a walk-through on an example problem, though.
  8. Jan 15, 2009 #7
  9. Jan 15, 2009 #8


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    If anyone is still following this thread: after checking with other students in the class, I found that the teacher had corrected the problem on the sheet in class - f(x) was supposed to equal 2x-x^3, which makes a major difference.

    NoMoreExams - I understand that you're trying to be helpful, and I assume that you must get people that don't bother searching elsewhere before posting, but I do. I searched Google, and had the page up for Tangent Lines on Wikipedia before you provided the Derivatives link. I had even run several advanced searches on Physics Forums to try to find very similar problems to mine.

    So even though I may be the exception, I find it frustrating when people tell me to check in other places when I, personally, use posting on a forum only as a last resort (I don't believe in swamping forums with repeats of previous topics).
  10. Jan 15, 2009 #9
    But doesn't the 2nd page on the google search I linked give you the necessary tutorial you need to solve your problem?
  11. Jan 15, 2009 #10


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    The tutorial in NoMoreExams' link does cover the process needed to solve your problem.
  12. Jan 16, 2009 #11
    Didnt you try this method? It'll give the answer straightaway,

    [tex]y=2x - x^3 [/tex]
    [tex]\frac{\delta y}{\delta x}=~ 2-3x^2[/tex]

    Now, this is the equation for the slope, substituting value of 'x' you'll get the slope of the line to be -10!
    Then have this value in the equation mentioned above to get the equation of the tangent!

    P.S:NoMoreExams please note this.:smile:
  13. Jan 16, 2009 #12
    What am I noting? That evaluating the derivative at a point will give you the slope of the tangent line? I kind of knew that already. My point was that the derivative is not the same as (y - y1)/(x - x1) as you wrote it.
  14. Jan 16, 2009 #13
    [tex] (y-y_1)=\frac{\delta y}{\delta x} (x-x_1) [/tex]

    This is the exact representation of the equation of the required equation!
  15. Jan 16, 2009 #14
    Once again, that's for a tangent line in which case it's true that the derivative of the tangent line will be its slope. That's what derivative, or one of the things, a derivative represents the slope of your curve at that point i.e. the slope of the tangent line to your curve at a particular point. My point wasn't to say you're flat out wrong, it was to clarify your point.
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