fcalc03_ppt_0303 by alison bacigalupo

3.3 Rules for Differentiation

Quick Review In Exercises 1-6, write the expression as a sum of powers of x. 1.

− 2 ) ( x −1 + 1)

2 5 3. 3 x − + 2 x x 2

−1

+ 2 ) ( x −2 + 1)

−1

 x  2.  2 ÷  x +1 

3x 4 − 2 x3 + 4 4. 2x2 x −1 + x −2 6. x −3

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Quick Review 7. Find the positive roots of the equation 2 x 3 − 5 x 2 − 2 x + 6 = 0 and evaluate the function y = 500 x 6 at each root. Round your answers to the nearest integer, but only in the final step.

8. If f ( x ) = 7 for all real numbers x, find

( a ) f ( 10 ) ( c)

f ( x + h)

( b ) f ( 0)

f ( x) − f ( a) ( d ) lim x →a x−a

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Quick Review 9. Find the derivatives of these functions with respect to x.

( a)

f ( x ) =π

( b)

f ( x ) =π 2

( c)

f ( x ) = π 15

10. Find the derivatives of these functions with respect to x using the definition of the derivative. x π a f x = b f x = ( ) ( ) ( ) ( ) π x

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Quick Review Solutions In Exercises 1-6, write the expression as a sum of powers of x. 1.

− 2 ) ( x −1 + 1)

x + x 2 − 2 x −1 − 2 2 5 3. 3 x − + 2 x x 2

3 x 2 − 2 x −1 + 5 x −2 5.

−1

+ 2 ) ( x + 1) −2

x −3 + x −1 + 2 x −2 + 2

−1

 x  2.  2 ÷  x +1  x + x −1

3x 4 − 2 x3 + 4 4. 2x2 3 2 x − x + 2 x −2 2 x −1 + x −2 6. −3 x x2 + x

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Quick Review Solutions 7. Find the positive roots of the equation 2 x 3 − 5 x 2 − 2 x + 6 = 0 and evaluate the function y = 500 x 6 at each root. Round your answers to the nearest integer, but only in the final step. At x =1.173, 500 x 6 ≈1305

At x = 2.394, 500 x 6 ≈ 94, 212

8. If f ( x ) = 7 for all real numbers x, find

( a ) f ( 10 )

( c) f ( x + h)

( b ) f ( 0)

7 7

( d ) lim x→a

f ( x) − f ( a) x−a

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Quick Review Solutions 9. Find the derivatives of these functions with respect to x.

( a)

f ( x) =π 0

( b)

f ( x) =π 2 0

( c)

f ( x ) = π 15 0

10. Find the derivatives of these functions with respect to x using the definition of the derivative. x π ( a ) f ( x) = ( b) f ( x) = π x 1 ′ f ( x) = f ′ ( x ) = − π x −2 π

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What you’ll learn about 

  

Positive Integer Powers, Multiples, Sums and Differences Products and Quotients Negative Integer Powers of x Second and Higher Order Derivatives

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Rule 1 Derivative of a Constant Function If f is the function with the constant value c, then df d = ( c) = 0 dx dx This means that the derivative of every constant function is the zero function.

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Rule 2 Power Rule for Positive Integer Powers of x. If n is a positive integer, then d n x ) = nx n −1 ( dx The Power Rule says: To differentiate x n , multiply by n and subtract 1 from the exponent.

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Rule 3 The Constant Multiple Rule If u is a differentiable function of x and c is a constant, then d du cu = c ( ) dx dx This says that if a differentiable function is multiplied by a constant, then its derivative is multiplied by the same constant.

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Rule 4 The Sum and Difference Rule If u and v are differentiable functions of x, then their sum and differences are differentiable at every point where u and v are differentiable. At such points, d du dv ( u ± v) = ± dx dx dx.

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Example Positive Integer Powers, Multiples, Sums, and Differences 3 Differentiate the polynomial y = x 4 + 2 x 2 − x − 19 4 dy That is, find . dx

By Rule 4 we can differentiate the polynomial term-by-term, applying Rules 1 through 3. dy d 4 d d 3  d = ( x ) + ( 2 x 2 ) −  x ÷− ( 19 ) dx dx dx dx  4  dx 3 = 4 x + 2 ×2 x − − 0 4 3 = 4 x3 + 4 x − 4 3

Sum and Difference Rule

Constant and Power Rules

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You Try Positive Integer Powers, Multiples, Sums, and Differences Find dy/dx x3 x 2 y = + + x + 100 3 2

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Example Positive Integer Powers, Multiples, Sums, and Differences Does the curve y = x 4 − 8 x 2 + 2 have any horizontal tangents? If so, where do they occur? Verify you result by graphing the function. If any horizontal tangents exist, they will occur where the slope is equal to zero. To find these points we will set Calculate

dy dx

dy = 0 and solve for x. dx

dy d 4 = ( x − 8 x 2 + 2 ) = 4 x3 − 16 x dx dx

dy Set = 0 and solve for x dx 4 x 3 − 16 x = 0

4 x ( x 2 − 4 ) = 0; 4 x = 0

x2 − 4 = 0

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Rule 5 The Product Rule The product of two differentiable functions u and v is differentiable, and d dv du uv = u + v ( ) dx dx dx The derivative of a product is actually the sum of two products.

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Example Using the Product Rule Find f ′ ( x ) if f ( x ) = ( x 3 − 4 ) ( x 2 + 3 ) Using the Product Rule with u = x3 − 4 and v = x 2 + 3, gives d f ′ ( x ) = ( x 3 − 4 ) ( x 2 + 3)  = ( x 3 − 4 ) 2 x + ( x 2 + 3) 3x 2 dx = 2 x 4 − 8 x + 3x 4 + 9 x 2 = 5x4 + 9 x2 − 8x

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You Try Product Rule Let y = ( x + 1)( x 2 + 1). Find dy/dx. (a) by applying the product rule dy d ( x 2 + 1) d ( x + 1) 2 = ( x + 1) + ( x + 1) = dx dx dx = ( x + 1)(2 x) + ( x 2 + 1)(1) = 2 x 2 + 2 x + x 2 + 1 = 3x 2 + 2 x + 1

(b) by multiplying the factors first and then differentiation. y = x3 + x 2 + x + 1

dy = 3x 2 + 2 x + 1 dx Copyright ÂŠ 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

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Rule 6 The Quotient Rule

u At a point where v ≠ 0, the quotient y = of two differentiable v functions is differentiable, and du dv v −u d u dx dx  ÷= 2 dx  v  v Since order is important in subtraction, be sure to set up the numerator of the Quotient rule correctly.

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Example Using the Quotient Rule Find f ′ ( x )

x3 − 4 if f ( x ) = 2 x +3

Using the Quotient Rule with u = x 3 − 4 and v = x 2 + 3, gives 3 2 2 3 d  ( x − 4 )  ( x + 3) 3 x − ( x − 4 ) 2 x = f ′( x) =  2 2 2 dx  ( x + 3)  ( x + 3)  

3x 4 + 9 x 2 − 2 x 4 + 8 x

+ 3)

x4 + 9 x2 + 8x

+ 3)

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You Try Using the Quotient Rule Find f ′ ( x )

( x − 1)( x 2 + x + 1) if f ( x ) = x3 Using the Quotient Rule with u = ( x − 1)( x 2 + x + 1) and v = x 3 ,

First we can expand ( x − 1)( x 2 + x + 1) = x 3 + x 2 + x − x 2 − x − 1 = x 3 − 2  ( x − 1) x + x + 1)  d  x 3 − 1  ( d f ′( x) =  =  x3  3 dx  x dx     d 3 3 d 3 3 x ( x − 1) − ( x − 1) (x ) ( ) dx dx = 2 ( x3 )

x ) ( 3x ) − ( x ( = 3

− 1) ( 3x 2 )

3x 5 − 3x 5 + 3x 2 = x3

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Example Using the Quotient Rule Find f ′ ( x )

x3 − 4 if f ( x ) = 2 x +3

Using the Quotient Rule with u = x 3 − 4 and v = x 2 + 3, gives 3 2 2 3 d  ( x − 4 )  ( x + 3) 3 x − ( x − 4 ) 2 x = f ′( x) =  2 2 2 dx  ( x + 3)  ( x + 3)  

3x 4 + 9 x 2 − 2 x 4 + 8 x

+ 3)

x4 + 9 x2 + 8x

+ 3)

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Example Working with Numerical Values Suppose u and v are functions of x that are differentiable at x = 0, and that u(0) = 5, u '(0) = −3, v(0) = −1, v '(0) = 2. d (uv ) (a) Find the value of at x = 0 dx d ( uv ) = u(0)v '(0) + v(0)u '(0) dx = (5)(2) + ( −1)( −3) = 10 + 3 = 13

(b) Find the values of

d v  ÷ dx  u 

d  v(0)  u (0)v '(0) − v(0)u '(0) = 2 dx  u (0) ÷ ( u (0))  =

(5)(2) − (−1)(−3) 10 − 3 7 = = 2 25 25 (5)

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You Try Working with Numerical Values Suppose u and v are functions of x that are differentiable at x = 0, and that u (0) = 5, u '(0) = −3, v(0) = −1, v '(0) = 2. Find the values of

d u  ÷ dx  v 

d  v(0)  v(0)u '(0) − u (0)v '(0) =  ÷ dx  u (0)  (v(0)) 2 =

( −1)( −3) − (5)(2) 3 − 10 = = −7 2 ( −1) 1

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Rule 7 Power Rule for Negative Integer Powers of x If n is a negative integer and x ≠ 0, then d n x ) = nx n −1 . ( dx This is basically the same as Rule 2 except now n is negative.

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Example Power Rule for Negative Integer Powers of x If n is a negative integer and x ≠ 0, then d n x ) = nx n −1 . ( dx This is basically the same as Rule 2 except now n is negative. d  x3 + x  Find dx  2 x 6 ÷  d  x3 + x  d  1 1  d  1 −3 1 −5  = + =  x + x ÷  3 5 ÷ dx  2 x 6 ÷ dx 2 x 2 x 2   dx  2   1 1 3 5 3 5 = ( −3) x −4 + ( −5) x −6 = − x −4 − x −6 = − 4 − 6 2 2 2 2 2x 2x 3x 2 + 5 =− 2 x6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

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You try Power Rule for Negative Integer Powers of x d  x4 + 1  Find dx  x 3 ÷  d  x4 + 1  d  1  d −3 = x + = x + x ( )  3 ÷ dx  x 3 ÷ dx x dx    =1 + ( −3) x −4 = 1 − 3 x −4 = 1 − 3x 2 + 5 =− 2 x6

3 x4

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Example Negative Integer Powers of x Find an equation for the line tangent to the curve y =

1 at the point ( 1,1) . x

Rewrite the function as y = x −1 and use the Power Rule to find the derivative. −1 y ′ = − 1x −2 = 2 1 x y= x −1 Evaluate y ′ ( 1) = = − 1 1 The line through ( 1,1) with slope m = − 1 is y − 1 = − 1( x − 1)

y=− x + 2

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You Try Negative Integer Powers of x 1 Find an equation for the line tangent to the curve y = 2 at the point 2x 1 −2 Rewrite the function as y = x and use the Power Rule to 2 find the derivative. 1 −1 y ′ = ( −2) x −3 = 3 2 x −1 Evaluate y ′ ( 1) = = − 1 1 The line through ( 1,1) with slope m = − 1 is y−

 1  1, ÷.  2

1 = − 1 ( x − 1) 2 3 y=− x + 2

1 ). 2

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Second and Higher Order Derivatives dy is called the first derivative of y with respect to x. dx The first derivative may itself be a differentiable function of x. If so, The derivative y ′ =

dy ′ d  dy  d 2 y its derivative, y ′′ = =  ÷= 2 , dx dx  dx  dx is called the second derivative of y with respect to x. If y ′′

( y double prime )

is differentiable, its derivative,

dy ′′′ d 3 y y ′′′ = = 3, dx dx is called the third derivative of y with respect to x.

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Second and Higher Order Derivatives The multiple-prime notation begins to lose its usefulness after three primes. d n â&#x2C6;&#x2019;1 n So we use y ( ) = y ( ) " y super n" dx to denote the nth derivative of y with respect to x. Do not confuse the notation y ( ) with the nth power of y, which is y n . n

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Example Second and Higher Order Derivatives y = 3x 6 − 2 x 5 + 2 x 2 + 1 dy y′ = = 18 x 5 − 10 x 4 + 4 x dx d2y y ′′ = 2 = 90 x 4 − 40 x 3 + 4 dx y (3)

d3y = 3 = 360 x 3 − 120 x 2 dx

y = x3 − x2 + 2 x + 1 dy y′ = = 3x 2 − 2 x + 2 dx d2y y ′′ = 2 = 6 x − 2 dx y (3) y (4)

d3y = 3 =6 dx d4y = 4 =0 dx Slide 3- 32

Quick Quiz Sections 3.1 – 3.3 You may use a graphing calculator to solve the following problems. 1. Let f ( x ) = x + 1 . Which of the following statements about f are true? I. f is continuous at x = − 1. II. f is differentiable at x = − 1. III. f has a corner at x = − 1.

( A) ( B) ( C) ( D)

I only II only III only I and III only

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( A) ( B) ( C) ( D)

I only II only III only I and III only

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Quick Quiz Sections 3.1 – 3.3 2. If the line normal to the graph of f at the point ( 1, 2 ) passes through

the point ( −1,1) , then which of the following gives the value of f ′ ( 1) = ?

( A) ( B)

−2 2

−1 2 1 ( D) 2 ( E) 3

( C)

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Quick Quiz Sections 3.1 – 3.3 2. If the line normal to the graph of f at the point ( 1, 2 ) passes through

the point ( −1,1) , then which of the following gives the value of f ′ ( 1) = ?

( A) ( B)

−2 2

−1 2 1 ( D) 2 ( E) 3

( C)

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Quick Quiz Sections 3.1 – 3.3 3.

dy 4x − 3 if y = . dx 2x +1

Find

( A)

( 4 x − 3)

( B) − ( C)

( 4 x − 3)

( 2 x + 1)

( D) − ( E)

( 2 x + 1)

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Quick Quiz Sections 3.1 – 3.3 3.

dy 4x − 3 if y = . dx 2x +1

Find

( A)

( 4 x − 3)

( B) − ( C)

( 4 x − 3)

( 2 x + 1)

( D) − ( E)

( 2 x + 1)

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