MATH 31 REVIEWS Chapter 2: Derivatives

Chapter 2: Derivatives “ If you don't go off on a tangent while studying, you can derive a great deal of success from it. ”

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) h0 h

f(x) = 3x2 - 4x + 1 f (x) = lim f(x+h) - f(x) h0 h = lim [3(x+h)2 - 4(x+h) + 1] - [3x2 - 4x + 1] h0 h

f(x) = 3x2 - 4x + 1 f (x) = lim f(x+h) - f(x) h0 h = lim [3(x+h)2 - 4(x+h) + 1] - [3x2 - 4x + 1] h0 h = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1

f (x) = 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 h0 h

f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1 = lim 6xh + 3h2 - 4h h0 h = lim h (6x + 3h - 4)

f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1 = lim 6xh + 3h2 - 4h h0 h = lim h (6x + 3h - 4) = lim (6x + 3h - 4) = 6x - 4 h0

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.

e.g. Sketch the derivative function y = f (x) y = f(x)

y = f(x) Horizontal slopes ...

y = f(x) Horizontal slopes ... y = f (x) ... become x-intercepts (f  = 0)

y = f(x) Slope > 0 y = f (x)

y = f(x) Slope > 0 y = f (x) Positive y-coordinates

y = f(x) Slope < 0 y = f (x) Negative y-coordinates

Degree = 3 y = f(x) y = f (x) 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)

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

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

y = -2x2 - 8x - (10x + 20 - 4x2 - 8x) (x2 + 4x)2 y = -2x2 - 8x - 10x - 20 + 4x2 + 8x y = 2x2 - 10x - 20 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) 2 dx 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) 2 dx y = 1 (4 + 3x - 9x2) -1/2  (3 - 18x) 2

y = 1 (4 + 3x - 9x2) -1/2  (3 - 18x) 2 y = 3 - 18x 2 4 + 3x - 9x2 or y = 3 (1 - 6x)

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 when 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 (x2) y + x2  y - 15x2 = 4y3  y + 0 Product rule Chain rule

x2y - 5x3 = y4 + 1 (x2) y + x2  y - 15x2 = 4y3  y + 0 2x y + x2 y - 15x2 = 4y3 y

x2y - 5x3 = y4 + 1 (x2) y + x2  y - 15x2 = 4y3  y + 0 2x y + x2 y - 15x2 = 4y3 y x2 y - 4y3 y = 15x2 - 2xy Get y terms on one side of the equations. All others go on the other side.

x2y - 5x3 = y4 + 1 (x2) y + x2  y - 15x2 = 4y3  y + 0 2x y + x2 y - 15x2 = 4y3 y x2 y - 4y3 y = 15x2 - 2xy y (x2 - 4y3) = 15x2 - 2xy Factor out the y

x2y - 5x3 = y4 + 1 (x2) y + x2  y - 15x2 = 4y3  y + 0 2x y + x2 y - 15x2 = 4y3 y x2 y - 4y3 y = 15x2 - 2xy y (x2 - 4y3) = 15x2 - 2xy y = 15x2 - 2xy x2 - 4y3