So this week in the wonderful world of Calculus BC: Taylor Polynomial and Approximations.
I would just like to say that I have absolutley no clue at all what these are or how you do them. So that is pretty much my question for the week.
After teaching Mu A practice I realized that I am kind of rusty with my derivatives.
So today's post will be about derivatives.
d/dx [uv] = uv' + vu'
d/dx [u/v] = (vu' - uv') / v^2
d/dx [sinu] = (cosu)(u')
d/dx [cosu] = -(sinu)(u')
d/dx [tanu] = (secu)^2(u')
d/dx [cotu] = -(cscu)^2(u')
d/dx [secu] = (secu)(tanu)(u')
d/dx [cscu] = -(cscu)(cotu)(u')
d/dx [ln(u)] = u' / u
d/dx [u] = (u)(u') / (u)
d/dx [e^u] = e^u * u'
One rule for derivative that the Mu A's weren't really getting was chain rule. I told that the way I remember it was to work from the outside in.
Peace Out,
Ryan
Sunday, November 7, 2010
Subscribe to:
Post Comments (Atom)
Pn(x) = f(c) + F'(c)(x-c) + (f''(c)(x-c)^2)/2! + ... (f^n(c)(x-c)^n)/n!
ReplyDeleteuse this formula, take derivative, plug in. DONE
Betway sign up bonus - JTM Hub
ReplyDeleteSign up using promo code 춘천 출장안마 BETWAY100 and get up to 양주 출장안마 £100 아산 출장마사지 in sports 구리 출장샵 bets with 충청북도 출장마사지 promo code 'EXTRABUNG'. Betway will reward you a deposit of up to