Definition of the Derivative help
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Definition of the Derivative help

[From: ] [author: ] [Date: 12-05-03] [Hit: ]
and your 5 lol ?!......

df(x)/dx = lim(e-->0){(f(x+e) - f(x))/e}

If you think about it, the term f(x+e) - f(x) is just the change in the value of f evaluated at two nearby points and e is the interval in x over which that change occurs so the fraction {f(x+e) - f(x)}/e is the slope of a straight line defined by the values of f and the interval in x of e. The limit function draws these two points togetherso the straight line approximation now becomes exact.

Let's work one simple example of a function as a derivative using the above definition. Let f(x) = x^2 and we''l find df/dx

df/dx = lim(e-->0){(x+e)^2 -x^2)/e} = lim(x-->0){(x^2 + 2ex + e^2 -x^2)/e}

= lm(e-->0){(2ex + e^2)/e} = lim(e-->0){2x + e} now take e = 0

df/dx = 2x

You can immediately see that if f(x) = c a constant, then df/dx = 0 as it has to be since f(x) = c will not change for any value of x.

You can appy this definition to any function of f(x) that is continuous and with some care, you can also find derivatives of discontinuous functions in the intervals were thaey are continuous. Fortunately, there are rules for doing differentiation so you don't always have to go through this limit process. for instance, if f(x) = x^n where n is some number then

df/dx = n*x^(n-1)

Now your f'(1) , etc requires you to first find df/dx and then evalute the derivative for x = 1 or whatever the numberical argument is. Your example:

f(x) = 2 - 3x^2

df/dx = -6x

f'(1) = -6, f'(2) = -12 etc.

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i'm not sure how to define it but the derivative is like taking a step forward you take your equation and take it to a new level
F(x)= -5 so the derivative of a constante number like 1 2 3 4 etc ... =0 so F'(x) = 0

F(x)=3x-7 let's go to the basic form it has the form of ax+b b is a normal number so =0 when you get the derivative so that leaves u with ax the derivative of that = a there's a bumch of rules u need to memorise in order to solve derivative questions so back to the equation F(x)=3x - 7 so F'(x) = 3 understood ?

F(x) = 2 - 3x^2 when u have the form x^n (n is a normal number in this case let's say it's 2) when u get the derivative of it it will become nx^n-1 in this case n= 2 that means 2x but here we have another number which is -3 so it's the form of ax^n that means the derivative would be anx^n-1
solve it derivative of 2 = 0 and the derivative of 3x^2 would be -3 x 2 X^2-1 because in this case a= -3 and n =2 so F'(x) = -6x also to find F'(1) f'(2) and F'(3) just replace x by the number between parantheses in the equation after u found the derivative F'(1) = -6(1) F'(2) = -6(2) F'(3)= -6(3)

did u understand what i explained if u need anything else feel free to ask ? and your 5 lol ?!
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