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§3.2 Some Differentiation Formulas

The student will learn about derivatives

of constants,

the product rule,notation,

of constants, powers, of constants, powers, sums and differences,

the quotient rule, and

the chain rule

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The Derivative of a Constant

Let y = f (x) = C be a constant function, then

y’ = f ’ (x) = 0.

What is the slope of a constant function?

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Example 1f (x) = 17

f ‘ (x) = 0

If y = f (x) = C then y’ = f ’ (x) = 0.

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Power Rule.

A function of the form f (x) = xn is called a power function. (Remember √x and all radical functions are power functions.)

Let y = f (x) = xn be a power function, then

y’ = f ’ (x) = n xn – 1.

THIS IS VERY IMPORTANT. IT WILL BE USED A LOT!

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Example 2f (x) = x5

f ‘ (x) = 5 • x4 = 5 x4

If y = f (x) = xn then y’ = f ’ (x) = n xn – 1.

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Example 3f (x) = 3 x

f (x) = , can be rewritten as f (x) = x1/3 and we can then find the derivative.

3 x

f ‘ (x) = 1/3 x - 2/3

f (x) = x 1/3

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Constant Multiple Property.

Let y = f (x) = k • u (x) be a constant k times a differential function u (x). Then

y’ = f ’ (x) = k • u’ (x) = k • u’.

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Example 4f (x) = 7x4

If y = f (x) = k • u (x) then f ’ (x) = k • u’.

f ‘ (x) = 7 • 28 x37 • 4 • x3 =

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Emphasisf (x) = 7x

If y = f (x) = k • u (x) then f ’ (x) = k • u’.

f ‘ (x) = 7 • 77 • 1 =

REMINDER: If f ( x ) = c x then f ‘ ( x ) = c

The derivative of x is 1.

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Sum and Difference Properties.

• The derivative of the sum of two differentiable functions is the sum of the derivatives. • The derivative of the difference of two differentiable functions is the difference of the derivatives.

OR

If y = f (x) = u (x) ± v (x), then

y ’ = f ’ (x) = u ’ (x) ± v ’ (x).

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Example 5

From the previous examples we get -

f (x) = 3x5 + x4 – 2x3 + 5x2 – 7 x + 4

f ‘ (x) = 15x4 + 4x3 – 6x2 + 10x – 7

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Example 6f (x) = 3x - 5 - x - 1 + x 5/7 + 5x- 3/5

f ‘ (x) = - 15x - 6 + x - 2 + 5/7 x – 2/7 - 3 x – 8/5

Show how to do fractions on a calculator.

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Notation

Given a function y = f ( x ), the following are all notations for the derivative.

y ′ f ′ ( x )

)x(fdx

d

dx

yd

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Graphing Calculators

Most graphing calculators have a built-in numerical differentiation routine that will approximate numerically the values of f ’ (x) for any given value of x.

Some graphing calculators have a built-in symbolic differentiation routine that will find an algebraic formula for the derivative, and then evaluate this formula at indicated values of x.

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Example 7

3. Do the above using a graphing calculator.

f (x) = x 2 – 3x and f ’ (x) = 2x - 3

Using dy/dx under the “calc” menu.

Let x = 2.

slope Tangent equation

Using tangent under the “draw” menu.

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Example 8 - TI-89 ONLY

Do the above using a graphing calculator with a symbolic differentiation routine.

f (x) = 2x – 3x2 and f ’ (x) = 2 – 6x

Using algebraic differentiation under the home “calc” menu.

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Median Summary.

If f (x) = C then f ’ (x) = 0.

If f (x) = xn then f ’ (x) = n xn – 1.

If f (x) = k • u (x) then f ’ (x) = k • u’ (x) = k • u’.

If f (x) = u (x) ± v (x), then

f ’ (x) = u’ (x) ± v’ (x).

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Derivates of ProductsThe derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Product Rule

)x('f)x(s)x('s)x(f])x(s)x(f[dx

d

OR 'fs'sf)sf(dx

d

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Example

Find the derivative of y = 5x2(x3 + 2).

Product Rule

Let f (x) = 5x2 then f ‘ (x) =

Let s (x) = x3 + 2 then s ‘ (x) =

= 15x4 + 10x4 + 20x = 25x4 + 20x

10x

3x2, and

)]x('f)x(s)x('s)x(f])x(s)x(f[dx

d

y ‘ (x) = 5x2 • 3x2 + (x3 + 2)y ‘ (x) = 5x2y ‘ (x) = 5x2 • 3x2y ‘ (x) = 5x2 • 3x2 + (x3 + 2) •10x

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Derivatives of Quotients The derivative of the quotient of two functions is the bottom function times the derivative of the top function minus the top function times the derivative of the bottom function, all over the bottom function squared.

Quotient Rule:

2])x(b[

)x('b)x(t)x('t)x(b

)x(b

)x(t

dx

d

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Derivatives of Quotients

May also be expressed as -

2b

'bt'tb

)x(b

)x(t

dx

d

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Example

Let t (x) = 3x and then t ‘ (x) =

Find the derivative of .5x2

x3y

Let b (x) = 2x + 5 and then b ‘ (x) =

2)5x2(

2x33)5x2()x('f 2)5x2(

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3.2.

2b

'bt'tb

)x(b

)x(t

dx

d

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Median Summary.

Product Rule. If f (x) and s (x), then

f • s ' + s • f ' sfdx

d

Quotient Rule. If t (x) and b (x), then

2b

'bt'tb

b

t

dx

d

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

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

The domain of m is the set of all numbers x such that x is in the domain of g and g (x) is in the domain of f.

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ExamplesLet f (u) = u4, g (x) = 2x + 5, and m (v) = ln v. Find:

f [ g (x)] =

g [ f (x)] = g (x4) =

m [ g (x)] =

f (2x + 5) = (2x + 5)4

m (2x + 5) =

2x 4 + 5

ln (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.

1nn xnxdx

d

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

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Chain Rule: Power Rule. General Power Rule. [Chain Rule]

Theorem 1. If u (x) is a differential function, n is any real number, and

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

then

f ’ (x) = n un – 1 u’or

dx

duunu

dx

d 1nn

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

I use u (x) because !!!

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Example 1Find the derivative of y = (x3 + 2) 5.

Let u (x) = x3 + 2, then y = u 5 and du/dx = 3x2

53 )2x(dx

d5 (x3 + 2) 3x24

= 15x2(x3 + 2)4

Chain Rule

dx

duunu

dx

d 1nn

NOTE: If we let u = x 3 + 2, then y = u 5.

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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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Example 2Find the derivative of y =

Rewrite as y = (x 3 + 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

= x2 (x3 + 3) –1/2

2

3

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Example 3

Find f ’ (x) if f (x) = .)8x3(

x2

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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

3 4

3

(3x 8)(4x ) 6xf '(x)

(3x 8)

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Summary.

Product Rule. If f (x) and s (x), then

f • s ' + s • f ' sfdx

d

Quotient Rule. If t (x) and b (x), then

2b

'bt'tb

b

t

dx

d

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Summary.

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

then

dx

duunu

dx

d 1nn

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ASSIGNMENT

§3.2: Page 52; 1 – 23 odd.