Lesson 2 Increment Method Ok Na Ok

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Lesson 2

Finding the Derivatives (Increment Lesson Objectives: At the end of the lesson, the students are expected to: 1. enumerate the steps in finding the derivatives of a function by increment method; 2. find the derivatives of the given function by applying the increment method; and 3. manifest the value of patience, perseverance and responsibility in doing one’s work.

Method) Since the derivative is defined as the limit of the difference quotient, the concept of limits can be applied in finding the derivatives. If for a fixed value of x, the quotient approaches to a limit as ∆x → 0, this limit is called the derivative of y with respect to x for a given value of y and is denoted by

dx . dy

Steps in finding the derivative of a function using the increment method: 1. Substitute all values of x’s in the given function y = f (x) by x + ∆x. 2. Subtract the given function from the values in step 1 and simplify. 3. Divide the result in step 2 by ∆x. 4. Take the result in step 3 as ∆x → 0.

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Illustrative Examples: 1. Find the derivative of y = 2x + 3. Step 1. Substitute all x’s by x + ∆x. y ‘ = 2 ( x + ∆x) + 3 Step 2. Subtract the given function from the values from step 1 and simplify. y ‘ = 2 ( x + ∆x) + 3 – ( 2x + 3) = 2x + 2∆x + 3 – 2x – 3 y ‘ = 2∆x Step 3. Divide the result in step 2 by ∆x. y ‘=

2∆x ∆x

y ‘ = 2. Step 4. Take the limit of the result for step 3 as ∆x → 0. Since there is no ∆x in the last equation in step 3, therefore, the derivative of y = 2x + 3 as ∆x → 0 is 2. 2. Find the derivative of y = 3x2 – 6x + 1. Step 1. Substitute all x’s by x + ∆x. y ‘ = 3 ( x + ∆x)2 – 6 (x + ∆x) + 1 Step 2. Subtract the given function from the values from step 1 and simplify. y ‘ = 3 (x + ∆x)2 – 6 (x + ∆x) + 1 – (3x2 – 6x + 1) = 3 (x2 + 2∆x2 + ∆2x2) – 6x – 6∆x + 1 – 3x2 + 6x – 1 = 3x2 + 6∆x2 + 3∆2x2 – 6x – 6∆x + 1 – 3x2 + 6x – 1 y ‘ = 6∆x2 + 3∆2x2 – 6∆x Step 3. Divide the result in step 2 by ∆x. ∆x 2 + 3∆2 x 2 − 6∆x y ‘= ∆x y ‘ = 6x + 3 ∆x - 6 Step 4. Take the limit of the result for step 3 as ∆x → 0.

234 Lim y ‘ = 6x + 3 ∆x – 6 ∆x →0

y ‘ = 6x – 6 3. Find the derivative of y =

3x + 5 . 2x − 4

Step 1. Substitute all x’s by x + ∆x. y‘= Simplifying, y’ =

3( x + ∆x) + 5 2( x + ∆x) − 4 3 x + 3∆x + 5 2 x + 2∆x − 4

Step 2. Subtract the given function from the values from step 1 and simplify. y‘=

3( x + ∆x) + 5 3x + 5 2( x + ∆x) − 4 2x − 4

Simplifying, y ‘ = (2x – 4) (3x + 3∆x + 5) – (3x + 5)(2x + 2∆x – 4) (2x + 2∆x – 4) (2x – 4) y‘=

22∆x (2 x + 2∆x − 4)(2 x − 4)

Step 3. Divide the result in step 2 by ∆x.

y‘=

22∆x 1 . (2 x + 2∆x − 4)(2 x − 4) ∆x

Step 4. Take the limit of the result for step 3 as ∆x → 0.

Lim y ‘ = ∆x →0

y‘ =

22 x (2 x + 2∆x − 4)(2 x − 4) 22 (2 x − 4)(2 x − 4)

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y‘ =

22 (2 x − 4) 2 Exercise 2 The Increment Method

Name: ________________________________ Date: Year and Section: ______________________ Score:

____________ ____________

Find the derivatives of the following functions using the increment method. 1. y = - 15

2. y = - x + 10

3. y = 4x – 12

4. y = 3x2 – 2x + 5

5. y = 6x2 + 10x – 3

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6. y =

2x + 5 3x − 2

7. y =

x2 +1 11x + 5

8. y =

1 x

9. y =

5x − 4

10. y =

2x − 5 3x

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