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Interpretation of the second derivative : Now that we've discover some higher order derivatives we have to probably talk regarding an interpretation of the second derivative.
If the position of an object is specified by s(t) we know that the velocity is first derivative of the position.
v (t ) = s′ (t )
First derivative of any velocity is the acceleration of object; however since it is the first derivative of the position function we can also think of the acceleration as the second derivative of the position function.
a (t ) = v′ (t ) = s′′ (t )
Alternate Notation : There is couple of alternate notation for higher order derivatives. Recall as well that there was a fractional notation for the first derivative.
f ′ ( x ) = df /dx
We should extend this to higher order derivatives.
f ′′ ( x )= d 2 y / dx f ′′′ ( x ) = d 3 y/ dx etc.
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Graph f ( x ) = e x and g ( x ) = e - x . Solution There actually isn't a lot to this problem other than ensuring that both of these exponentials are graphed somewhere.
1+1=
How do you find the distributive property any faster?
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