The Product & Quotient Rules - Solutions

October 18, 2017 | Author: wolfretonmaths | Category: Rates, Analysis, Abstract Algebra, Differential Calculus, Mathematical Analysis

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AQA Core 3 Differentiation Section 2: The product and quotient rules Solutions to Exercise 1. (i)

y  x(2 x  1)2 Let u  x

Use the chain rule here

du 1 dx

dv  2(2 x  1)  2  4(2 x  1) dx dy dv du Using the product rule: u v dx dx dx  x  4(2 x  1)  1 (2 x  1)2 Let v  (2 x  1)2

 4 x(2 x  1)  (2 x  1)2

(2x – 1) is a common factor

 (2 x  1)(4 x  (2 x  1))  (2 x  1)(6 x  1) (ii) y  x(2 x  1)2  x(4 x 2  4 x  1)  4 x 3  4 x 2  x

dy dx

 12 x 2  8 x  1

(iii) (2 x  1)(6 x  1)  12 x 2  8 x  1 so the two answers are algebraically equivalent.

2. y  x(2 x  1)4 du Let u  x  1 dx

Use the chain rule here

dv  4(2 x  1)3  2  8(2 x  1)3 dx dy dv du u v Using the product rule: dx dx dx  x  8(2 x  1)3  1 (2 x  1)4 Let v  (2 x  1)4

 8 x(2 x  1)3  (2 x  1)4  (2 x  1) (8 x  (2 x  1)) 3

(2x + 1)³ is a common factor

 (2 x  1)3(10 x  1)

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AQA C3 Differentiation 2 Exercise solutions 1

3. y  x 1  2 x  x(1  2 x )2 Use the chain rule here du Let u  x  1 dx 1 1 dv 1 1 Let v  (1  2 x )2   2 (1  2 x ) 2  2  dx 1  2x dy dv du Using the product rule: u v dx dx dx 1  x  1 1  2x 1  2x

x

 1  2x 1  2x x  (1  2 x )  1  2x 1  3x  1  2x

4.

Use a common denominator to write this as a single fraction

1 3

y  x 3(1  x )

du Let u  x 3   3x 2 dx 2 2 1 dv 1 Let v  (1  x )3   3 (1  x ) 3  1  31 (1  x ) 3 dx dy dv du u v Using the product rule: dx dx dx

Use the chain rule here

 x 3  31 (1  x ) 3  3 x 2 (1  x )3 2

1

 31 x 2(1  x ) 3 ( x  9(1  x )) 2

1 3

2

x 2 (1  x) 3 is a

common factor

 31 x 2(1  x ) 3 (10 x  9) 2

5.

1 2

y  x 1  x  x(1  x ) du 1 Let u  x  dx dv 1 1  2 (1  x )  1   Let v  (1  x )  dx 2 1x dy dv du u v Using the product rule: dx dx dx 1 2

Use the chain rule here

1 2

 x  

1  1 x 1 2 1x

x 2 1x

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AQA C3 Differentiation 2 Exercise solutions At turning point,

x

 1x 2 1x x  2(1  x ) x  2  2x 3x  2

x

2 3

2

1

When x  23 , y  23 (1  23 )2 

3 3

2 2  Turning point is  , . 3 3 3

6. y 

x 2x  1

du 1 dx dv Let v  2 x  1  2 dx Let u  x

Using the quotient rule,

dy dx

v

du dv u dx dx

v2 1(2 x  1)  2 x  (2 x  1)2 

1 (2 x  1)2

1 is always negative for x  21 , 2 (2 x  1) and so the gradient of the curve is always negative for x  21 . Since (2 x  1)2 is always positive, 

7. y 

x2 1 x

du   2x Let u  x dx 1 1 1 dv 1  2 (1  x ) 2  1  21 (1  x ) 2 Let v  (1  x )2  dx 2

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Use the chain rule here

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AQA C3 Differentiation 2 Exercise solutions

Using the quotient rule,

dy dx

v

du dv u dx dx

v2

2 x(1  x )2  x 2  21 (1  x ) 2  1x 4 x(1  x )  x 2  3 2(1  x )2 4 x  3 x 2 x(4  3 x )  3  3 2(1  x )2 2(1  x )2 1

8. y 

1

Multiply top and 1

bottom by 2(1  x) 2

x2 x 1

du  2x dx dv Let v  x  1  1 dx Let u  x 2

Using the quotient rule,

At turning points,

dy dx

v

du dv u dx dx

v2 2 x( x  1)  x 2  ( x  1)2 x( x  2)  ( x  1)2

x( x  2) 0 ( x  1)2 x  0 or x  2

When x = 0, y = 0

22 4 21 The turning points are (0, 0) and (2, 4). When x = 2, y 

9. y 

1 x 1 x 1

Let u  1  x 2 1

Let v  1  x 2

du 1  21 2x dx 1 dv    21 x  2 dx 

Using the quotient rule,

dy dx

v

du dv u dx dx

v2

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AQA C3 Differentiation 2 Exercise solutions x  (1  x )  (  21 x  )(1  x )  (1  x )2  1 2 x (1  x  1  x )  (1  x )2 x  (1  x )2 1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 x (1  x )2

10. (i)

w 

1 x 1x

Let u  1  x Let v  1  x

du 1 dx dv   1 dx

Using the quotient rule:

v

dw  dx

du dv u dx dx

v2 (1  x )  (1  x )( 1)  (1  x )2 

2 (1  x )2

3

1 x  3 (ii) y    w 1  x   dy  3w 3 dw Using the chain rule:

dy dx

dy dw

dw dx

 3w 2 

2 (1  x )2 2

2 1 x   3   2  1  x  (1  x ) 6(1  x )2  (1  x )4

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