Thread: Limits
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Old 09-16-2003, 07:56 AM   #12 (permalink)
Sapper
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Some very good answers here.

Change f(x)= 3x^2 + 4x + 2 into "box"

Then plug "box" into the differentiation formula in place of "x".

you get: (box - h) - box / h

now simplify the terms using the real value of "x" (ie. 3x^2 + 4x + 2)

Later on, you will learn "the power rule" which simply states:

where f(x)= x ^ 3
f ' (x)= [exp] x ^ [exp - 1]

ie. f ' (x) = 3x^2

.. much better

In your case:

f(x)=3x^2 + 4x + 2

f ' (x)=6x + 4x + 0

PS: you will notice that the derivitave of any "real number" is always 0
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