objectives: 1. be able to find the derivative of function by applying the chain rule critical...
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Objectives:1. Be able to find the derivative of function by applying the Chain Rule
Critical Vocabulary:Chain Rule
Warm Ups:1. Find the derivative of f(x) = (3x - 5)2 using the power rule.2. Find the derivative of f(x) = (3x - 5)2 using the product rule.
Warm Ups:1. Find the derivative of f(x) = (3x - 5)2 using the power rule.
f(x) = (3x - 5)(3x - 5)
f(x) = 9x2 - 30x + 25
f’(x) = 18x - 30
2. Find the derivative of f(x) = (3x - 5)2 using the product rule.
f(x) = (3x - 5)(3x - 5)
g(x) = 3x - 5 g’(x) = 3 h(x) = 3x - 5 h’(x) = 3
f’(x) = (3x - 5)(3) + (3x - 5)(3)
f’(x) = 9x – 15 + 9x - 15
f’(x) = 18x – 30
The Chain Rule: The chain rule deals with the composition of functions.
Without the Chain Rule With the Chain Rule
1)( 2 xxf 1)( 2 xxf
23)( xxf 5)23()( xxf
2)( xxf 3 2)( xxf
They are like onions, they have layers
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x and
dx
du
du
dy
dx
dy )('))(('))](([ xgxgfxgf
dx
d
Composition Function
Idea: If y = un then y’ = n·un-1 · u’
Onion Outside
Inside
If y = (3x + 2)5 then y’ = 5(3x + 2)4 · 3
un
u5
n un-1 u’
5 u4 3
Example 1: f(x) = (3x - 5)2
h(u) = u2 h’(u) = 2u
g(x) = 3x - 5 g(x) = 3
f’(x) = 2(3x - 5) • 3
f’(x) = 6(3x - 5)
f’(x) = 18x - 30
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x and
dx
du
du
dy
dx
dy )('))(('))](([ xgxgfxgf
dx
d
Example 2: f(x) = (x2 + 1)3
h(u) = u3 h’(u) = 3u2
g(x) = x2 + 1 g(x) = 2x
f’(x) = 3(x2 + 1)2 • 2x
f’(x) = 6x(x2 + 1)2
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x and
dx
du
du
dy
dx
dy )('))(('))](([ xgxgfxgf
dx
d
Example 3: 3 22 3)( xxf 32
2 3)( xxf
3
2
)( uuh 3
1
3
2)('
uuh
3)( 2 xxg xxg 2)('
xxxf 233
2)(' 3
12
31
2 33
4)('
xxxf
31
2 33
4)('
x
xxf
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x and
dx
du
du
dy
dx
dy )('))(('))](([ xgxgfxgf
dx
d
3 2 33
4)('
x
xxf
33
34)('
2
3 22
x
xxxf
Example 4: Find the equation of the tangent line when x = 2
32
234)(
xxf
3
2
)(
uuh 3
5
3
2)('
uuh
234)( xxg xxg 6)('
)6(343
2)(' 3
52 xxxf
35
2344)('
xxxf
35
234
4)('
x
xxf
First lets find the y-value
32
2)2(34)2(
f
32
8)2(
f
4
1)2( f
35
2)2(34
)2(4)2('
f
32
8)2('
f
4
1)2(' f
f(x) = mx + b
b )2(4
1
4
1
b2
1
4
1
b4
3
4
3
4
1)( xxf
3 222 3434
4)('
xx
xxf
22
3 2
34
344)('
x
xxxf
Page 290 #7-23 odds
Example 5: 22 1)( xxxf 2)( xxg xxg 2)('
2
12 )1()( xxh )2()1(
2
1)(' 2
12 xxxh
2
12 )1()('
xxxh
21
22
122 12)1()(' xxxxxxf
21
22
123 12)1()(' xxxxxf
222
12 12)1()(' xxxxxf
222
12 22)1()(' xxxxxf
23)1()(' 22
12
xxxxf
2
12
2
)1(
)23()('
x
xxxf
2
3
1
23)('
x
xxxf
Product Property
Distribute x2
Factor GCF
Distribute 2
Combine Like Terms
No Negative Exponents
2
223
1
1213)('
x
xxxxxf
Example 6: 3 2 4)(
x
xxf
xxg )( 1)(' xg
31
2 4)( xxh xxxh 243
1)(' 3
22
32
2 43
2)('
xxxh
32
22 4)]([ xxg
32
2
3
223
12
4
432
14
)('
x
xxxx
xf
32
2
3
2223
12
4
43
24
)('
x
xxxxf
32
2
223
22
4
3
244
)('
x
xxxxf
32
2
23
22
4
43
14
)('
x
xxxf
32
2
2
3
22
4
3
124
)('
x
xx
xf
34
2
2
4
3
12
)('
x
x
xf
34
2
2
43
12)('
x
xxf
3 22
2
443
12)('
xx
xxf
22
3 223 222
43
4124)('
x
xxxxf
Example 7: Find the equation of the tangent line at (2, 2) 53
1)( 2 xxxf
xxg3
1)(
3
1)(' xg
21
2 5)( xxh 21
2 5)('
xxxh
21
22
12 5
3
15
3
1)('
xxxxxf
21
22
122 5
3
15
3
1)('
xxxxf
)5(53
1)(' 222
12
xxxxf
5253
1)(' 22
12
xxxf
53
52)('
2
2
x
xxf
Derivative
5)2(3
5)2(2)2('
2
2
f
9
13)2(' f Slope
b
29
132
b9
262
b9
8
9
8
9
13)( xxf
Equation of Tangent Line
53
5552)('
2
222
x
xxxxf
Page 290 - 291 #25 - 35 odds