I looked it up in the dictionary and still can't figure it out!

You've heard the expression: Fools rush in where angels fear to tread? Well with Dr. Gerard, Witeoutking, Maiku, Diane 302, available, What am I doing here? I don't have good sense and you need some kind of an answer concerningderivatives

AW,

Calculus is a section/category of Math just like Algebra and Geometry are. Only it is learnt if you pursue Math at a higher level.

You must be familliar with graphs, right? Okay...
The value for the gradient (also known as slope)is the 'differentiated' value. For e.g. in a distance-time graph, if you differntiate the value of the graph
the measurement obtained (at a certain point or x-co ordinate) gives you the speed.

Differentiation is usually used in curved graphs.
Satisfied?

Another way to approach this is to start with a very simple, real-world example, and explain the derivative in (hopefully) a more intuitive fashion. Suppose you throw a baseball straight up in the air at an initial speed of 20 m/s (meters/second). On a vertical axis, we'll call the baseball's original position (where it leaves your hand) 0 meters. The baseball rises to a height of about 20 meters, then falls back to your hand (assuming you catch it), with a height of 0 meters again. This whole process takes about 4 seconds.

Neglecting air resistance, we can describe the baseball's height with the following equation:

y = (20 m/s) t - (4.9 m/s²) t² ,

where y is the height in meters above your hand, and t is the time in seconds after you begin the toss. Notice that in terms of units, y is in meters.

One of the most common applications of the derivative in physics is to take the derivative with respect to time. In this example, the starting equation gives an object's position at any given time, and the units are in meters. If we take the derivative with respect to time (t), then we get a new equation with different units:

y′ = 20 m/s - (9.8 m/s²) t

The new equation gives the object's velocity (instead of position) at any given time and the units are in meters/second (instead of meters). Notice that the new units have seconds in the denominator. Every time you take the derivative with respect to time, you divide the units of your answer by your time unit (seconds). (Note that positive velocity represents upward movement and negative velocity represents downward movement.)

If we take the derivative again, then we divide the units of our answer by our time unit (seconds) again, to get meters/second², which is acceleration. So the new equation gives the object's acceleration at any given time:

y′′ = - 9.8 m/s²

This, of course is just the constant negative (downward) acceleration of gravity.

So just remember, every time you take the derivative with respect to a variable, you're dividing your answer by that variable's units. So when you take the derivative with respect to time, you get a natural progression from your original function (in this case position), to the rate of change of the original function (in this case position/time, or velocity), to the secondary rate of change or how fast the rate of change is changing (in this case position/time², or acceleration).

This message has been edited. Last edited by: silenteuphony,