Spm Addmath_differentiation Exercise Ii

ONE-SCHOOL.NET Differentiation Exercise II Sum and Difference Rule

5. y = 2 x −

y =u±v u and v are functions in x dy du dv = ± dx dx dx

2 x

Example y = 2 x3 + 5 x 2 dy = 2(3) x 2 + 5(2) x = 6 x 2 + 10 x dx Exercise 1 dy Find of the following dx 1. y = 3 x 2 − x

6. y =

2 2 x− 2 5 x

7. y =

3 1 − 5 3 5x x

2. y = 120 x 3 − 21x + 2

2

2 3. y = −3 x 3 − x − 1 3

4. y = 9 x 4 − 12 x 3 − 2

1

ONE-SCHOOL.NET 8. y =

Chain Rule

3x3 1 − 2 2 2x

y = un

u and v are functions in x

dy dy du = × dx du dx

Example y = (2 x 2 + 3)5 u = 2 x 2 + 3, y = u5 ,

therefore

therefore

du = 4x dx

dy = 5u 4 du

dy dy du = × dx du dx = 5u 4 × 4 x = 5(2 x 2 + 3) 4 × 4 x = 20 x(2 x 2 + 3) 4

9. y =

3x3 − 2 x + 4 x

Or differentiate directly y = (2 x 2 + 3)5 dy = 5(2 x 2 + 3) 4 × 4 x = 20 x(2 x 2 + 3) 4 dx Exercise 2 Differentiate the following equation upon x. 1. y = (2 x + 3)3

10. y =

5 x3 − 7 x + 4 2x2

2. y = (2 x 2 + 3x + 3) 2

2

ONE-SCHOOL.NET 2 3. y = ( + 3x 3 )3 x

6. y =

1 2x + 3

5 2 3 + ) x 2 x3

7. y =

5 (5 x + 3x) 2

8. y =

6 4(7 x + 3 x − 4) 2

4. y = (

5. y = 3(4 x + 3 x ) 4

2 4

3

3

2

ONE-SCHOOL.NET 9. y =

Product Rule

2 −(2 x − 7) 2

2

y = uv u and v are functions in x dy du dv =v +u dx dx dx Example y = (2 x + 3)(3 x 3 − 2 x 2 − x) u = 2x + 3 v = 3x3 − 2 x 2 − x du dv =2 = 9 x2 − 4x − 1 dx dx dy du dv =v +u dx dx dx 3 2 =(3x − 2 x − x)(2) + (2 x + 3)(9 x 2 − 4 x − 1) Or differentiate directly y = (2 x + 3)(3 x 3 − 2 x 2 − x) dy = (3 x 3 − 2 x 2 − x)(2) + (2 x + 3)(9 x 2 − 4 x − 1) dx

10. y =

Exercise 3 Find f’(x) of the following equation 1. f ( x) = x 2 (2 x + 3)3

−3 3 8(9 x + 21x)8

4

ONE-SCHOOL.NET 2.

f ( x) = x 5 (2 x 3 + 3 x) 2

5.

f ( x) = (7 x + 6) 2 (3x 2 − 12) 4

3.

f ( x) =

x (5 x + 3 x 2 ) −3 5

6.

f ( x) = (5 x + 1)3 x 4

4.

f ( x) = ( x 2 − 12 x)(5 x + 11) 2

5

ONE-SCHOOL.NET Quotient Rule

u y= v

u and v are functions in x

2. y =

2 x2 − 2 3x − 1

3. y =

2x 2 x − 3x

du dv v −u dy = dx 2 dx dx v Example x2 y= 2x +1 u = x2 v = 2x +1 du dv = 2x =2 dx dx du dv v −u dy = dx 2 dx dx v dy (2 x + 1)(2 x) − x 2 (2) = dx (2 x + 1) 2 =

4 x2 + 2 x − 2 x2 2 x2 + 2 x = (2 x + 1) 2 (2 x + 1) 2

Or differentiate directly x2 y= 2x +1 dy (2 x + 1)(2 x) − x 2 (2) = (2 x + 1) 2 dx =

4 x2 + 2 x − 2 x2 2 x2 + 2 x = (2 x + 1) 2 (2 x + 1) 2

Exercise 4 Find the first derivative of the following equation x +1 1. y = x −1

6

3

ONE-SCHOOL.NET 4. y =

5. y =

x +1 3x − 3x 2

1 − 2x 2( x 3 − 3)

7

6. y =

2 − 3x 5( x 2 − 3x)

7. y =

( x − 1)3 2 − 3x

ONE-SCHOOL.NET 8. y =

2x2 + 3 (3x − 12) 2

9. y =

7x +1 5 − 2 x + 12 4 x + 1

10. y = (

8

x 3 ) 2x −1