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10.5 Basic Differentiation Properties. Instead of finding the limit of the different quotient to obtain the derivative of a function, we can use the rules.

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Presentation on theme: "10.5 Basic Differentiation Properties. Instead of finding the limit of the different quotient to obtain the derivative of a function, we can use the rules."— Presentation transcript:

1 10.5 Basic Differentiation Properties

2 Instead of finding the limit of the different quotient to obtain the derivative of a function, we can use the rules of differentiation (shortcuts to find the derivatives).

3 Review: Derivative Notation There are several widely used symbols to represent the derivative. Given y = f (x), the derivative may be represented by any of the following: ■ f ’(x) ■ y’ ■ dy/dx

4 The graph of f (x) = C is a horizontal line with slope 0, so we would expect f ’(x) = 0. Theorem 1. Let y = f (x) = C be a constant function, then y’ = f ’(x) = 0. What is the slope of a constant function?

5 Examples: Find the derivatives of the following functions 1)f(x) = -24 f’(x) = ? 2) f(x) = f’(x) = ? 3) f(x) = π f’(x) = ? 0 0 0

6 Power Rule A function of the form f (x) = x n is called a power function. This includes f (x) = x (where n = 1) and radical functions (fractional n). Theorem 2. (Power Rule) Let y = x n be a power function, then y’ = f ’(x) = n x n – 1.

7 Example Differentiate f (x) = x 10. Solution: By the power rule, the derivative of x n is n x n–1. In our case n = 10, so we get f ’(x) = 10 x 10-1 = 10 x 9

8 Differentiate Solution: Rewrite f (x) as a power function, and apply the power rule: Example

9 Examples: Find the derivatives of the following functions 1)f(x) = x 6 f’(x) = ? 2) f(x) = t -2 f’(x) = ? 3) f(x) = t 3/2 f’(x) = ? 4) f(x) = f’(x) = ? 5) f(x) = f’(x) =? 6) f(x) = f’(x) =? 6x 5 -2t -3 = = x -1 = u 2/3 = x -1/2

10 Constant Multiple Property Theorem 3. Let y = f (x) = k  u(x) be a constant k times a function u(x). Then y’ = f ’(x) = k  u’(x). In words: The derivative of a constant times a function is the constant times the derivative of the function.

11 Example Differentiate f (x) = 10x 3. Solution: Apply the constant multiple property and the power rule. f ’(x) = 10  (3x 2 ) = 30 x 2

12 Examples: Find the derivatives of the following functions 1)f(x) = 4x 5 f’(x) = ? 2) f(x) = f’(x) = ? 4) f(x) = f’(x) = 5) f(x) = f’(x) = 20x 4

13 Sum and Difference Properties Theorem 5. If y = f (x) = u(x) ± v(x), then y’ = f ’(x) = u’(x) ± v’(x). In words: ■ 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.

14 Differentiate f (x) = 3x 5 + x 4 – 2x 3 + 5x 2 – 7x + 4. Solution: Apply the sum and difference rules, as well as the constant multiple property and the power rule. f ’(x) = 15x 4 + 4x 3 – 6x 2 + 10x – 7. Example

15 Examples: Find the derivatives of the following functions 1)f(x) = 3x 4 -2x 3 +x 2 -5x+7 f’(x) = ? 2) f(x) = 3 - 7x -2 f’(x) = ? 4) f(x) = f’(x) =? 5) f(x) = f’(x) =?

16 Applications Remember that the derivative gives the instantaneous rate of change of the function with respect to x. That might be: ■ Instantaneous velocity. ■ Tangent line slope at a point on the curve of the function. ■ Marginal Cost. If C(x) is the cost function, that is, the total cost of producing x items, then C’(x) approximates the cost of producing one more item at a production level of x items. C’(x) is called the marginal cost.

17 Example: application Let f (x) = x 4 - 8x 3 + 7 (a) Find f ’(x) (b) Find the equation of the tangent line at x = 1 (c) Find the values of x where the tangent line is horizontal Solution: a)f ’(x) = 4x 3 - 24x 2 b)Slope: f ’(1) = 4(1) - 24(1) = -20 Point: (1, 0) y = mx + b 0 = -20(1) + b 20 = b So the equation is y = -20x + 20 c) Since a horizontal line has the slope of 0, we must solve f’(x)=0 for x 4x 3 - 24x 2 = 0 4x 2 (x – 6) = 0 So x = 0 or x = 6

18 Example: more application The total cost (in dollars) of producing x radios per day is C(x) = 1000 + 100x – 0.5x 2 for 0 ≤ x ≤ 100. 1.Find the marginal cost at a production level of radios. Solution: C’(x) = 100 – x 2) Find the marginal cost at a production level of 80 radios and interpret the result. Solution: The marginal cost is C’(80) = 100 – 80 = $20 It costs $20 to produce the next radio (the 81 st radio) 3)Find the actual cost of producing the 81st radio and compare this with the marginal cost. Solution: The actual cost of the 81st radio will be C(81) – C(80) = $5819.50 – $5800 = $19.50.

19 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  x n then f ’ (x) = k  n x n-1 If f (x) = u(x) ± v(x), then f ’(x) = u’(x) ± v’(x).


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