3.6 Notes.notebook
September 24, 2013
3.6 Chain Rule
Name: __________________
Objectives: Students will be able to differentiate composite
functions using the chain rule and find slopes of parametrized
curves.
We're experts at differentiating functions like f(x) = x2 - 4, but
how do we differentiate a composite function like g(x) = sin(x2 - 4)?
GIVE UP? CRY? RUN AWAY?
NO WAY! We use the CHAIN RULE!
THE CHAIN RULE
If f is differentiable at the point u = g(x), and g is differentiable at
x, then the composite function (f o g)(x) = f(g(x)) is differentiable
at x, and _______________________.
In Leibniz notation, if y = f(u) and u = g(x), then
_______________________,
where dy/du is evaluated at u = g(x).
Sep 10­4:04 PM
Examples
1.) Use the given substitution and the Chain Rule to find dy/dx.
y = tan(2x - x2), u = 2x - x2
2.) An object moves along the x-axis so that its position at any time
t ≥ 0 is given by x(t) = s(t). Find the velocity of the object as a
function of t.
s = sin[(3π/2)t] + cos[(7π/4)t]
Sep 10­9:01 PM
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3.6 Notes.notebook
September 24, 2013
Power Chain Rule If u is a differentiable function of x, then we
can extend the chain rule:
Examples
1.) y = (x + 2)50
2.) y = -2(6x + 2)10
3.) y = (x2 + 5)100
Sep 13­4:45 PM
Find dy/dx.
4.) y = (1 + cos2x)3
5.) y =
x
√1 + x2
6.) Find dr/dθ. r = sec2θtan2θ
Sep 10­9:08 PM
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3.6 Notes.notebook
September 24, 2013
1.) Find the value of (f o g)' at the given value of x.
f(u) = 1 - 1/u , u = g(x) = 1/(1-x), x = -1
2.) Find the equation of the line tangent to the curve at the point
t = -1/6 with x = sin2πt and y = cos2πt.
Sep 10­9:14 PM
Suppose that the functions f and g and their derivatives have the
following values at x = 2 and x = 3.
x
f(x)
g(x)
f'(x)
g'(x)
2
8
2
1/3
-3
3
3
-4
2π
5
Evaluate the derivatives with respect to x of the following
combinations at the given value of x.
1.) 3g(x) at x = 3
2.) g(x)/f(x) at x = 2
3.) (f3(x) - 2g(x))5 at x = 3
Sep 13­7:05 PM
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3.6 Notes.notebook
September 24, 2013
Homework: pages 153-155: 1-39 odd, 41, 43, 53, 55, 56, 58, 63
Sep 13­7:07 PM
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