§2.3 the chain rule and higher order derivatives

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§2.3 The Chain Rule and Higher Order Derivatives

The student will learn aboutcomposite functions,

the chain rule, and

nondifferentiable functions.

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Composite Functions Definition. A function m is a composite of functions f and g if

m (x) = f ◦ g = f [ g (x)]

This means that x is substituted into g first. The result of that substitution is then substituted into the function f for your final answer.

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ExamplesLet f (u) = u 3 , g (x) = 2x + 5, and m (v) = │v│. Find: f [ g (x)] =

g [ f (x)] = g (x3) =

m [ g (x)] =

f (2x + 5) = (2x + 5)3

m (2x + 5) =

2x 3 + 5

│2x + 5│

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Chain Rule: Power Rule. We have already made extensive use of the power rule with xn,

We wish to generalize this rule to cover [u (x)]n, where u (x) is a composite function. That is it is fairly complicated. It is not just x.

1nn xnxdxd

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Chain Rule: Power Rule. That is, we already know how to find the derivative of

f (x) = x 5

We now want to find the derivative of

f (x) = (3x 2 + 2x + 1) 5

What do you think that might be?

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General Power Rule. [Chain Rule]

If u (x) is a function, n is any real number, and

If f (x) = [u (x)]n

thenf ’ (x) = n un – 1 u’

orn n 1d u nu

xx dddu

* * * * * VERY IMPORTANT * * * * *

The chain

Chain Rule: Power Rule.

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ExampleFind the derivative of y = (x3 + 2) 5.

Let the ugly function be u (x) = x3 + 2. Then

53 )2x(dxd

5 (x3 + 2) 3x24

= 15x2(x3 + 2)4

n n 1d u nuxx dd

du

Chain Rule

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ExamplesFind the derivative of:

y = (x + 3) 2

y = 2 (x3 + 3) – 4

y = (4 – 2x 5) 7 y’ = 7 (4 – 2x 5) 6 (- 10x 4)

y’ = 2 (x + 3) (1) = 2 (x + 3)

y’ = - 8 (x3 + 3) – 5 (3x 2)

y’ = - 70x 4 (4 – 2x 5) 6

y’ = - 24x 2 (x3 + 3) – 5

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ExampleFind the derivative of y =

Rewrite as y = (x 3 + 3) 1/2

= 3/2 x2 (x3 + 3) –1/2

3x 3

Then y’ = 1/2Then y’ = 1/2 (x 3 + 3) – 1/2Then y’ = 1/2 (x 3 + 3) – 1/2 (3x2)

Try y = (3x 2 - 7) - 3/2

y’ = (- 3/2) (3x 2 - 7) - 5/2 (6x)

= (- 9x) (3x 2 - 7) - 5/2

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ExampleFind f ’ (x) if f (x) = .

)8x3(x

2

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We will use a combination of the quotient rule and the chain rule.Let the top be t (x) = x4, then t ‘ (x) = 4x3

Let the bottom be b (x) = (3x – 8)2, then using the chain rule b ‘ (x) = 2 (3x – 8) 3 = 6 (3x – 8)

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432

))8x3(()8x3(6x)x4()8x3()x('f

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Remember Def: The instantaneous rate of change for a function, y = f (x), at x = a is:

This is the derivative.

h)a(f)ha(flim

0h

Sometimes this limit does not exist. When that occurs the function is said to be nondifferentiable.

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Remember Def: The instantaneous rate of change for a function, y = f (x), at x = a is:

This is the derivative and a graphing way to represent the derivative is as the slope of the curve. This means that at some points on some curves the slope is not defined.

h)a(f)ha(flim

0h

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“Corner point”

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Vertical Tangent

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Discontinuous Function

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"If a function f …"

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Summary.

Ify = f (x) = [u (x)]n

then

dxduunu

dxd 1nn

Nondifferentiable functions.

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ASSIGNMENT§2.3 on my website

11, 12, 13.