On the absolute power thing, just make the number positive and raise it to that power. Like your question, abs[3]^3 is the same as 3^3, which is 27. If it was abs[-3]^3 then you would make it positive and have 3^3 again.
It's right. The derivative of what you typed is
[math]2xsin(x)+x^2cos(x)+2cos(x)-2xsin(x)-2cos(x)[/math]
Then everything but [math]x^{2}cos(x)[/math] cancles out.
On the downside to all who couldn't get this puzzle, I remember doing this back in elementry school. I couldn't get it then, but a lot of other people did. Maybe now that we're all so much smarter we think to much into it.
Okay, there aren't many steps to show for this one. You just have to realize that the derivative of Sec(x) is Sec(x)Tan(x), so your answer is just Sec(x)+C
I thought every AP Calc test had to be taken today so nobody could give out questions illegally (since every test is the same)? And what do you mean what section? If you were referring to my previous post that's for the whole test
For the first one, just foil it out and rewrite [math]cot^{2}(x)[/math] as [math]\frac{cos^{2}(x)}{sin^{2}(x)}[/math]. You get [math]sin^{2}(x)+cos^{2}(x)=1[/math] which is a common identity.
For the second, rewrite it as [math]\frac{sin(x)}{cos(x)}+\frac{cos(x)}{sin(x)}[/math]. After you cross multiply and add them together, you get [math]\frac{sin^{2}(x)+cos^{2}(x)}{cos(x)sin(x)}[/math]. The top is equal to 1 (the identity in your first problem), and when you break apart the bottom and rewrite it it is [math]sec(x)csc(x)[/math]
On the scoring guide I have its 0-24 is a 1, 25-39 is a 2, 40-57 is a 3, 58-74 is a 4, and 75-108 is a 5. Each free response is worth 9 points (totaling 54), and for multiple choice you do [number correct - (1/4 * number guessed wrong)]*1.2, meaning there is a max 54 points.
I finished everything with a little time to spare, but I really hate it when as soon as I finish and can't go back I figure something out and realize I did it wrong. I only left one question blank on the whole test
I'm taking it tomorrow too, AB. What year are you ecoli? I'm 11th. I'm a little nervous, but I've done really well on all the practice tests I've taken so far so I should be fine. Good luck to you
(This came right from The Simpsons)
What's so funny about the derivative of [math]\frac{r^{3}}{3}[/math]?
It's [math]r^{2}dr[/math] or r*dr*r (hardy-har-har) ... lame I know.
Like matt grime said, it depends on the function. You can check a few things though. If you want it to be differentiable at every real x then it has to be continous at every real x. Also, make sure there are no sharp points, like in the absolute value function at 0.
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