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Deep understanding of derivatives

  1. Jan 17, 2015 #1
    Hello everyone,i want to know about derivatives in detail!!suppose a function say a=x^2 has derivative 2x,i want to know what does that mean?how we'll prove it?if we put x=2,then a=4 and if we put x=3,we'll get a=9,does that mean a=9-4=5,the change?
    Sorry for the long question..
    Thanks!!
     
  2. jcsd
  3. Jan 17, 2015 #2

    HallsofIvy

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    The derivative is the "instantaneous rate of change". Think about why Newton and Leibniz created the Calculus. They were both, as were most scientists of the time, interested in what kept the planets in their orbits. They knew "acceleration= mass times force", and believed that the force depended on the distance from the sun, but that made no sense with their current definitions of "speed" and "acceleration". At any given instant, a planet has a specific distance from the sun so a specific force is being applied to it so "mass times force" has a specific value at a given instant. But speed is "distance moved divided by the time required". It is impossible to talk about "speed at a give instant"! Acceleration is "change in speed divided by the time required" so is even worse! How can you have a quantity, acceleration, which cannot have a value at a given instant, be equal to "mass times force" which does?

    As I said, the derivative is the "instantaneous rate of change" so we can define "instantaneous" speed as the derivative of the distance with respect to time and define "instantaneous" acceleration as the derivative of the speed with respect to time.

    For example, suppose we are looking at an object moving along the parabola [itex]y= x^2[/itex] such that at any time "t", x= 4t and [itex]y= 16t^2[tex]. At time t= 1, we are at (4, 16) and at time t= 2 we are at (8, 64), Our average velocity between t= 1 and t= 2 would be the vector <(8-4)/1, (64- 16)/1>= <4, 48>. Our average velocity between t= 1 and t= 1.5 (at time t= 1.5, our position is [itex]4(1.5), 16(1.5)^2)= (6, 36)[/itex] would be the vector <(6- 1)/.5, (36- 16)/.5>= <10, 40>. But using that definition of "velocity" as "change in position divided by change in time" we could not even talk about "velocity at t= 1" since there would be no "change in time" to divide by. But with Calculus giving "instantaneous rate of change, we can say that the instantaneous velocity at any time, t, is <4, 32t> so at t= 1 the instantaneous speed is < 4, 32>.

    A slightly more "geometric" way of putting this is that, as long as something is moving with constant speed, v, its position from some give "starting point" is d= vt+ d0. If you were to graph that, with t on the horizontal axis, d on the vertical, then the graph is a straight line with slope0 v. If the speed is not constant, the graph is not a straight line and there is no "slope". But, as long as the graph is "smooth", it has a tangent line at each point and we can use the slope of the tangent line at a given point as "the derivative" at that point.
     
  4. Jan 18, 2015 #3

    Svein

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    You just introduce a change - first you calculated for x= 2 and the for x= 3. Thus, the mean rate of change from x=2 to x=3 is (9 - 4)/(3 - 2) = 5. When you are trying to find a derivative, you keep your reference point (say x= 2) and then calculate the mean rate of change from x=2 to x=2.5, from x=2 to x=2.1 etc. Doing it this way, you get a sequence of numbers. If this sequence has a limit, that is the value of the derivative of the function at your reference point.

    As an aside, in physics we try to stay away from derivatives as much as possible. They are OK in theory, but if your "function" consists of a set of measurements, those measurements will contain noise and measurement errors. A derivative of something with noise in it will tend to show the noise, not the real "function".
     
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