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AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

Tuesday, January 24, 2006


Another scribe that I can't seem to find the time to do.

Well here it goes.

For the past weeks we have learned to differentiate many equations. Ugly or not so ugly, we can differentiate it. But now we run the rules backwards.

Who would've thought that a derivative and an integral undo what each other does? They are both very different from one another. Unlike addition and subtraction or multiplication and division, its not so obvious that they are inverses of one another. But they are, they undo what the other does.

Keep your bank of derivatives in hand, you'll need it. But this time add a couple of new rules, for antidifferentiating. Don't forget your plus C's, why? Because the derivative of any constant is 0. A big fat ZERO. The difference now when you're antidifferentiating is that you're not only getting ONE particular function but a WHOLE FAMILY of functions.

Advice for the entire chapter:
1) Keep your bank of derivative. You may add new ones...
2) Make a bank of anti-derivatives.
3) + C, never forget that.


Natural Log function and Rational function in particular 1/x


Input ---> Output
its inverse is...
Output ---> Input


X (domain) ---> Y (range)
its inverse is...
Y(range) ----> X (domain)


So where am I trying to lead this to?

If we look at the rational function 1/x
its domain is the reals and x does not equal to 0.

its anti-derivative should have the same domain.

This stands true, antiderivative of 1/x is ln x+c for x>0.

How about when x<0?

Using chain rule.
d/dx ln(-x) = (-1)1/(-x)= 1/x

So therefore, the antiderivative of 1/x is ln absolute(x) + C.

Don't forget that absolute sign or the + C as they have great distinctions to your answer.


Sorry for the late post.

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