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Could anyone link me to tutorial sheets with the most common derivative forming rules or state some common rules I need to follow when forming derivatives. (Is that what you call the below process?) Example would be

Dy/dx = x^2 + y^2

D^2y/dx^2 = 2x + 2y dy/dx

D^3/dx^3 = 2 + 2yd^2/dx^2 + 2(dy/dx)^2

(Here I know that the derivative is found by using
The First Term x Derivative of second term + Second term x Derivative of First .

I hope this makes sense. I cant find a better way to put it :( .
Example :

2007-05-02 03:40:44 · 3 answers · asked by Biju M 2 in Science & Mathematics Mathematics

3 answers

We usually call this "taking" the derivative. In my references, I have linked to a table of common derivatives and to some rules for taking more complicated functions. Your textbook probably has similar information in an appendix. It sounds like you understand the basic principal, so you should be able to use this information.

2007-05-02 03:44:52 · answer #1 · answered by DavidK93 7 · 0 0

differentiation is the rate of change... you've got the basics but its quite confusing!

I'll use the analogy of a car on a journey, you have the function of a curve that can give you any point on the curve (this would be your speed of the car at any given point);

y = x^3 + 5x + 3

the first derivative give you the rate of change of x with respect to y

d^2y/dx^2 = 3x^2 + 5

this gives you the tangent of the curve (and would indicate acceleration of your car)

The second derivative would give you the rate of change of y with respect to x of the rate of change of y with respect to x;

d^3y/dx^3 = 6x

This gives you the tangent of the tangent of the rate of change (this would show you if your acceleration is accelerating)

I know its a bit confusing but i believe if you understand what your doing mathematically then you'll find it a bit easier. Hope that helps ^_^

2007-05-02 10:56:54 · answer #2 · answered by dark_massiah 3 · 0 0

I refer to this page occasionally

http://www.sosmath.com/tables/derivative/derivative.html

2007-05-02 11:01:06 · answer #3 · answered by Kathleen K 7 · 0 0

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