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1 §3.2 Some Differentiation Formulas The student will learn about derivatives of constants, the product rule,notation, of constants, powers,of constants,

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Presentation on theme: "1 §3.2 Some Differentiation Formulas The student will learn about derivatives of constants, the product rule,notation, of constants, powers,of constants,"— Presentation transcript:

1 1 §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

2 2 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?

3 3 Example 1 f (x) = 17 f ‘ (x) = 0 If y = f (x) = C then y’ = f ’ (x) = 0.

4 4 Power Rule. A function of the form f (x) = x n is called a power function. (Remember √x and all radical functions are power functions.) Let y = f (x) = x n be a power function, then y’ = f ’ (x) = n x n – 1. THIS IS VERY IMPORTANT. IT WILL BE USED A LOT!

5 5 Example 2 f (x) = x 5 f ‘ (x) = 5 x 4 =5 x 4 If y = f (x) = x n then y’ = f ’ (x) = n x n – 1.

6 6 Example 3 f (x) = f (x) =, can be rewritten as f (x) = x 1/3 and we can then find the derivative. f ‘ (x) =1/3 x - 2/3 f (x) = x 1/3

7 7 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’.

8 8 Example 4 f (x) = 7x 4 If y = f (x) = k u (x) then f ’ (x) = k u’. f ‘ (x) = 7 28 x 3 7 4 x 3 =

9 9 Emphasis f (x) = 7x If y = f (x) = k u (x) then f ’ (x) = k u’. f ‘ (x) = 7 7 7 1 = REMINDER: If f ( x ) = c x then f ‘ ( x ) = c The derivative of x is 1.

10 10 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).

11 11 Example 5 From the previous examples we get - f (x) = 3x 5 + x 4 – 2x 3 + 5x 2 – 7 x + 4 f ‘ (x) =15x 4 + 4x 3 – 6x 2 + 10x– 7

12 12 Example 6 f (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.

13 13 Notation Given a function y = f ( x ), the following are all notations for the derivative. y ′f ′ ( x )

14 14 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.

15 15 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. slopeTangent equation Using tangent under the “draw” menu.

16 16 Example 8 - TI-89 ONLY Do the above using a graphing calculator with a symbolic differentiation routine. f (x) = 2x – 3x 2 and f ’ (x) = 2 – 6x Using algebraic differentiation under the home “calc” menu.

17 17 Median Summary. If f (x) = C then f ’ (x) = 0. If f (x) = x n then f ’ (x) = n x n – 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).

18 18 Derivates of Products The 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 OR

19 19 Example Find the derivative of y = 5x 2 (x 3 + 2). Product Rule Let f (x) = 5x 2 then f ‘ (x) = Let s (x) = x 3 + 2 then s ‘ (x) = = 15x 4 + 10x 4 + 20x =25x 4 + 20x 10x 3x 2, and y ‘ (x) = 5x 2 3x 2 + (x 3 + 2) y ‘ (x) = 5x 2 y ‘ (x) = 5x 2 3x 2 y ‘ (x) = 5x 2 3x 2 + (x 3 + 2) 10x

20 20 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:

21 21 Derivatives of Quotients May also be expressed as -

22 22 Example Let t (x) = 3x and then t ‘ (x) = Find the derivative of. Let b (x) = 2x + 5 and then b ‘ (x) = 3. 2.

23 23 Median Summary. Product Rule. If f (x) and s (x), then f s ' + s f ' Quotient Rule. If t (x) and b (x), then

24 24 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.

25 25 Examples Let f (u) = u 4, g (x) = 2x + 5, and m (v) = ln v. Find: f [ g (x)] = g [ f (x)] =g (x 4 ) = m [ g (x)] = f (2x + 5) =(2x + 5) 4 m (2x + 5) = 2x 4 + 5 ln (2x + 5)

26 26 Chain Rule: Power Rule. We have already made extensive use of the power rule with x n, We wish to generalize this rule to cover [u (x)] n. 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

27 27 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 u n – 1 u’ or * * * * * VERY IMPORTANT * * * * * I use u (x) because !!!

28 28 Example 1 Find the derivative of y = (x 3 + 2) 5. Let u (x) = x 3 + 2, then y = u 5 and du/dx =3x 2 5(x 3 + 2)3x 24 = 15x 2 (x 3 + 2) 4 Chain Rule NOTE: If we let u = x 3 + 2, then y = u 5.

29 29 Examples Find the derivative of: y = (x + 3) 2 y = 2 (x 3 + 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 (x 3 + 3) – 5 (3x 2 ) y’ = - 70x 4 (4 – 2x 5 ) 6 y’ = - 24x 2 (x 3 + 3) – 5

30 30 Example 2 Find the derivative of y = Rewrite as y = (x 3 + 3) 1/2 Then y’ = 1/2Then y’ = 1/2 (x 3 + 3) – 1/2 Then y’ = 1/2 (x 3 + 3) – 1/2 (3x 2 ) Try y = (3x 2 - 7) - 3/2 y’ = (- 3/2) (3x 2 - 7) - 5/2 (6x) = (- 9x) (3x 2 - 7) - 5/2 = x 2 (x 3 + 3) –1/2

31 31 Example 3 Find f ’ (x) if f (x) = We will use a combination of the quotient rule and the chain rule. Let the top be t (x) = x 4, then t ‘ (x) =4x 3 Let the bottom be b (x) = (3x – 8) 2, then using the chain rule b ‘ (x) = 2 (3x – 8) 3 =6 (3x – 8)

32 32 Summary. Product Rule. If f (x) and s (x), then f s ' + s f ' Quotient Rule. If t (x) and b (x), then

33 33 Summary. If y = f (x) = [u (x)] n then

34 34 ASSIGNMENT §3.2: Page 52; 1 – 23 odd.


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