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hey, how's it going?
Hi i really need this answer:)
take the derivative of (sin(3-x))^2 like it was x^2, you just have to remember to multiply by the derivative of what's inside the "squared"
oo sick thats so helpful thanks a million
hah wats the answer just to confirm?
why dont you give me what you came up with and we'll see if it looks good
so i know it starts of as 2(sin3-x)
then multiply by the derivative of sin(3-x)
so the derivative of sin(3-x) is? uhm cos(3-x)? correct?
not quite, remember you have to multiply by the derivative of what's inside the sin
waah i hate calc. ok so the derivative of that is 1?
lol idk im asking u!
what's the derivative of 3-x, you probably know that
so then its 2sin(3-x)(cos3-x)(1)
what'd we just talk about though?
usually you wouldn't leave it as times 1 just so you know, that would just go away
f(x)=sin^2(3-x) f'(x)= 2sin(3-x)(cos (3-x))(-1) Evaluate for 0: f'(0) = -2sin(3-(0))(cos(3-(0)) f'(0) = -2sin(3)cos(3) f'(0) = 0.279415..
Hi, Mr. Euclid. He forgot the -1, I have no intention of stealing your thunder ;)
no i don't care, and yeah i was already telling him but he disappeared
if you want some math to do here is a test for you http://floridamao.org/PublicPages/TestArchive.aspx?qy=true&category=States&style=Limits+%26+Derivatives&year=2005
Ooh cool, saved to bookmarks. I really should be doing my area between two curves calc homework, funny that I choose to do other math as an effective self distraction instead of more traditional routes (video games and food). Firefox hates this site, trying it with Chrome now. Love the name by the way. Euclid's Fifth Postulate and hyperbolic geometry ftw.