MATH 31 REVIEWS Chapter 2: Derivatives. “If you don't go off on a tangent while studying, you...

MATH 31 REVIEWS

Chapter 2: Derivatives

“ If you don't go off on a tangent while studying,

you can derive a great deal of success from it. ”

Chapter 2: Derivatives

1. Finding Derivatives from First Principles

Ex. If f(x) = 3x2 - 4x + 1 , find f (x) using limits.

f(x) = 3x2 - 4x + 1

f (x) = lim f(x+h) - f(x)

f(x) = 3x2 - 4x + 1

= lim [3(x+h)2 - 4(x+h) + 1] - [3x2 - 4x + 1]

f(x) = 3x2 - 4x + 1

= lim [3(x+h)2 - 4(x+h) + 1] - [3x2 - 4x + 1]

= lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1

f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1

= lim 6xh + 3h2 - 4h

f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1

= lim 6xh + 3h2 - 4h

= lim h (6x + 3h - 4)

f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1

= lim 6xh + 3h2 - 4h

= lim h (6x + 3h - 4)

= lim (6x + 3h - 4) = 6x - 4

2. Sketching Derivative Functions

Remember, the derivative represents the tangent slope of a function.

Thus, the derivative function simply describes the slopes

of the original function.

Sketch the derivative function y = f (x)

y = f(x)

y = f(x)Horizontal slopes ...

y = f (x)

... become x-intercepts (f = 0)

y = f(x)Slope > 0

Slope > 0

y = f (x)

y = f(x)Slope > 0

Slope > 0

y = f (x)Positive y-coordinates

Positive y-coordinates

y = f(x)

Slope < 0

y = f (x)

Negative y-coordinates

y = f(x)

y = f (x)

Degree = 3

Degree = 2

For polynomial functions, the degree decreases by 1

3. Power Rule

Differentiate y = 5x3 - 7x + 9 - 4 + 1 - 2

x x2 x3

y = 5x3 - 7x + 9 - 4 + 1 - 2

x x2 x3

y = 5x3 - 7x + 9 - 4x-1 + x-2 - 2x-3

Bring powers to the "top"

y = 5x3 - 7x + 9 - 4 + 1 - 2

x x2 x3

y = 5x3 - 7x + 9 - 4x-1 + x-2 - 2x-3

y = 5 (3x2) - 7 + 0 - 4 (-1x-2) + -2x-3 - 2(-3x-4)

Ignore the coefficients

Differentiate the powers

y = 5x3 - 7x + 9 - 4 + 1 - 2

x x2 x3

y = 5x3 - 7x + 9 - 4x-1 + x-2 - 2x-3

y = 5 (3x2) - 7 + 0 - 4 (-1x-2) + -2x-3 - 2(-3x-4)

y = 15x2 - 7 + 4x-2 - 2x-3 + 6x-4

y = 5x3 - 7x + 9 - 4 + 1 - 2

x x2 x3

y = 5x3 - 7x + 9 - 4x-1 + x-2 - 2x-3

y = 5 (3x2) - 7 + 0 - 4 (-1x-2) + -2x-3 - 2(-3x-4)

y = 15x2 - 7 + 4x-2 - 2x-3 + 6x-4

y = 15x2 - 7 + 4 - 2 + 6

x2 x3 x4

4. Product Rule

Differentiate y = (4x3 + 9x) (7x4 - 11x2 - 3) using

the product rule. No need to simplify the answer.

y = (4x3 + 9x) (7x4 - 11x2 - 3)

y = (4x3 + 9x) (7x4 - 11x2 - 3) + (4x3 + 9x) (7x4 - 11x2 - 3)

Product Rule: f g + f g

y = (4x3 + 9x) (7x4 - 11x2 - 3)

y = (4x3 + 9x) (7x4 - 11x2 - 3) + (4x3 + 9x) (7x4 - 11x2 - 3)

y = (12x2 + 9) (7x4 - 11x2 - 3) + (4x3 + 9x) (28x3 - 22x)

5. Quotient Rule

Differentiate y = 5 - 2x using the quotient rule.

x2 + 4x

y = 5 - 2x

x2 + 4x

y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x) (x2 + 4x)2

Quotient Rule: f g f g g2

y = 5 - 2x

x2 + 4x

y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x) (x2 + 4x)2

y = (-2) (x2 + 4x) - (5 - 2x) (2x + 4)

(x2 + 4x)2

y = 5 - 2x

x2 + 4x

y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x) (x2 + 4x)2

y = (-2) (x2 + 4x) - (5 - 2x) (2x + 4)

(x2 + 4x)2

y = -2x2 - 8x - (10x + 20 - 4x2 - 8x)

(x2 + 4x)2

y = -2x2 - 8x - (10x + 20 - 4x2 - 8x)

(x2 + 4x)2

y = -2x2 - 8x - 10x - 20 + 4x2 + 8x

(x2 + 4x)2

y = -2x2 - 8x - (10x + 20 - 4x2 - 8x)

(x2 + 4x)2

y = -2x2 - 8x - 10x - 20 + 4x2 + 8x

(x2 + 4x)2

y = 2x2 - 10x - 20

(x2 + 4x)2

or y = 2(x2 - 5x - 10)

x2 (x + 4)2

6. Chain Rule

Differentiate y = 4 + 3x - 9x2

y = 4 + 3x - 9x2

y = (4 + 3x - 9x2) 1/2

y = 4 + 3x - 9x2

y = (4 + 3x - 9x2) 1/2

y = 1 (4 + 3x - 9x2) -1/2 d (4 + 3x - 9x2)

Derivative of the "outside function"

Don't forget to find the derivative of the "inside function"

y = 4 + 3x - 9x2

y = (4 + 3x - 9x2) 1/2

y = 1 (4 + 3x - 9x2) -1/2 d (4 + 3x - 9x2)

y = 1 (4 + 3x - 9x2) -1/2 (3 - 18x)

y = 3 - 18x

2 4 + 3x - 9x2

y = 3 (1 - 6x)

2 4 + 3x - 9x2

7. Combination of Rules

Where does the function y = (2x - 7)4 (3x + 1)6

have a horizontal tangent?

y = (2x - 7)4 (3x + 1)6

y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]

Use the product rule first

y = (2x - 7)4 (3x + 1)6

y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]

y = 4(2x - 7)3 (2) (3x + 1)6 + (2x - 7)4 6(3x + 1)5 (3)

Use the chain rule next

Don't forget to do the derivative of the "inside function"

y = (2x - 7)4 (3x + 1)6

y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]

y = 4(2x - 7)3 (2) (3x + 1)6 + (2x - 7)4 6(3x + 1)5 (3)

= 8 (2x - 7)3 (3x + 1)6 + 18 (2x - 7)4 (3x + 1)5

Put both terms into the same "order"

y = (2x - 7)4 (3x + 1)6

y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]

y = 4(2x - 7)3 (2) (3x + 1)6 + (2x - 7)4 6(3x + 1)5 (3)

= 8 (2x - 7)3 (3x + 1)6 + 18 (2x - 7)4 (3x + 1)5

= 2 (2x - 7)3 (3x + 1)5 [ 4(3x + 1) + 9(2x - 7) ]

Factor out the common factors

y = (2x - 7)4 (3x + 1)6

y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]

y = 4(2x - 7)3 (2) (3x + 1)6 + (2x - 7)4 6(3x + 1)5 (3)

= 8 (2x - 7)3 (3x + 1)6 + 18 (2x - 7)4 (3x + 1)5

= 2 (2x - 7)3 (3x + 1)5 [ 4(3x + 1) + 9(2x - 7) ]

= 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]

y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]

When does the function have a horizontal tangent?

y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]

The function has a horizontal tangent when y = 0

2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ] = 0

y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]

The function has a horizontal tangent when y = 0

2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ] = 0

2x - 7 = 0 or 3x + 1 = 0 or 30x - 32 = 0

x = 7 x = -1 x = 16

2 3 15

8. Implicit Differentiation

Find y for the implicit relation x2y - 5x3 = y4 + 1

7. Implicit Differentiation

Find y for the implicit relation x2y - 5x3 = y4 + 1

Remember, y is a function of x. So, you must treat it like

a separate function f(x).

x2y - 5x3 = y4 + 1