4 November 2008
Hello everyone! Welcome to the first blog post of the second quarter!
Today we went over the Chain Rule.
Defined as: the derivative of the outside with respect to the inside multiplied by the derivative of the inside.
You can solve for the derivative using two methods.
For example:
f(x)=(3x+2) 3
using the inside/outside method…
f’(x)=3(3x+2)2 * 3
f’(x)=9(3x+2)2
substitution
u=3x+2 u’=3
y = u3
y’=3u2 * u1
y’=3(3x+2)2 * 3
now things get a little more complicated! take note, in the example give, there are essentially three functions: a cubic function, a sine function, and a binomial.
y=sin3(3x+4)
y’=3sin2(3x+4) * cos(3x+4) * 3
y’=9sin2(3x+4)cos(3x+4)
We also reviewed the Product Rule…
y=u*v
y’=u’v+uv’
let’s say we have y=x7(2x+5)10…
u=x7 u’=7×6
v=(2x+5)10 v’=10(2x+5)9 * 2
v’=20(2x+5)9
y’=u’v+uv’
y’= 7×6(2x+5)10 + 20×7(2x+5)9
We then looked at the quotient rule. Take note, because this is simply a multiplication of reciprocals, it will not be essential, but it will help immensely with certain problems.
Quotient Rule:
if y=u/v
then y’=(u’v-uv’)/v2
lets apply this…
y=(sin5x)/(8x-3)
u’=5cos5x
v’=8
y’= {(5cos(5x) * (8x-3)) – (8sin5x)}/(8x-3)2
homework
p.138 1–27 odd (not due thursday, we have a test)
and study for the big test this thursday
it will be on chapter 1 through 3 and some pre-calc. study trig and logs!
keep in mind the calculus graded challenge problem is due tuesday, nov 18
and multiple choice will be due on friday the 14th
