Basic Skills in Higher Mathematics

Basic Skills in

Higher Mathematics

Robert GlenAdviser in Mathematics

Mathematics 1(H)Outcome 3

Mathematics 1(Higher)

Outcome 2 Use basic differentiation

Differentiation

ff ((xx))dydydxdx



PC Index

ff ((xx))dydydxdx

PC(a) Basic differentiation

PC(b) Gradient of a tangent

PC(c) Stationary points

Click on the PC you want



PC(a) Differentiate a function reducible to a sum of powers of x

dydydxdx

ff ((xx))PC(a) - Basic

differentiation




PC(a) - Basic differentiation

1 Simple functions

2 Simple functions multiplied by a constant

3 Negative indices 4 Fractional indices

6 Sums of functions (simple cases)

7 Sums of functions (negative indices)

Click on the section you want

8 Sums of functions ( algebraic fractions)

5 Negative and fractional indices with constant




ff ((xx))dydydxdx


1 Simple functions




Every function f(x) has a related function called the derived function.The derived function is written f (x) “f dash x”

The derived function is also called the derivative.

To find the derivative of a functionyou differentiate the function.

Some examples

f(x)

x3

x6

x

3

3x2

6x5

1

0

f (x)

x10 10x9

(= x1)

(= 3x0)




Rule No. 1 for differentiation

If f(x) = xn , then f (x) = nxn -1




Differentiate each of these functions.

1 f(x) = x4

2 f(x) = x5

3 g(x) = x8

4 h(x) = x2

5 f(x) = x12

6 f(x) = x

Here are the answers

1 f (x) = 4x3

2 f (x) = 5x4

3 g(x) = 8x7

4 h(x) = 2x

5 f (x) = 12x11

6 f (x) = 1

7 f(x) = 5 7 f (x) = 0




Continue with Section 2

End of Section 1




ff ((xx))dydydxdx


2 Simple functions

multiplied by a constant




More examples

f(x)

2x3

3x6

2x10

5x

2 3x2

2 10x9

3 6x5

5 1

6x2

18x5

20x9

5

f (x)





If f(x) = axn , then f (x) = anxn -1





1 f(x) = 3x4

2 f(x) = 2x5

3 g(x) = ½ x8

4 h(x) = 5x2

5 f(x) = ¼ x12

6 f(x) = 8x


1 f (x) = 12x3

2 f (x) = 10x4

3 g(x) = 4x7

4 h(x) = 10x

5 f (x) = 3x11

6 f (x) = 8

7 f(x) = 10 6 f (x) = 0





End of Section 2




ff ((xx))dydydxdx


3 Negative indices









More examples

1 f(x) = 3

1x

= x -3

f (x) = -3 -4 -3 -2 -1 0-5

-1

-3-4

Note:This is an example of using Rule No.1with a negative index.

= 4

3x

x -4x ?




More examples

2 f(x) = x1

= x -1

f (x) = -1 -2 -1 0

-1

-1-2


= 2

1x

x -2x ?





1 f(x) = x -2

2 f(x) = x -4

3 g(x) =

4 h(x) =

5 f(x) =


1 f (x) = -2x -3

2 f (x) = -4x -5

3 g(x) = -5x -6 =

4 h(x) = -1x -2 =

5 f (x) = -10x -11 =

5

1x

x1

10

1x

6

5x

2

1x

11

10x





End of Section 3




ff ((xx))dydydxdx


4 Fractional indices









More examples

3 f(x) = 21

x

f (x) =

-1 0 1

-1

Note:This is an example of using Rule No.1with a fractional index.

= 21

2

1

x

21 2

1

21

21xx ?




More examples

4 f(x) = 32

x

f (x) =

-1 0 1

-1


= 31

3

2

x

32 3

2

31

31

xx ?





1 f(x) =

2 f(x) =

3 g(x) =

4 h(x) =


1 f (x) =

2 f (x) =

3 g(x) =

4 h(x) =

43

x

52

x

31

x 32

31

x

41

x 43

41 x

41

43 x

53

52

x





End of Section 4




ff ((xx))dydydxdx


5 Negative and fractional indices with constant





If f(x) = axn , then f (x) = anxn -1




More examples

5 f(x) = 3

2x

= 2x -3

f (x) = -6 -4 -3 -2 -1 0-5

-1

-3-4


= 4

6x

x -4x ?




More examples

6 f(x) = x5

= 5x -1

f (x) =-2 -1 0

-1

-1-2


= 2

5x

x -2x ?-5





1 f(x) = 3x -2

2 f(x) = 5x -4

3 g(x) =

4 h(x) =

5 f(x) =


1 f (x) = -6x -3

2 f (x) = -20x -5

3 g(x) = -10x -6 =

4 h(x) = -2x -2 =

5 f (x) = -30x -11 =

5

2x

x2

10

3x

6

10x

2

2x

11

30x




More examples

7 f(x) = 21

6x

f (x) =

-1 0 1

-1


= 21

3

x

21

21

21xx ? 6 ½ = 33




More examples

8 f(x) = 32

9x

f (x) =

-1 0 1

-1


= 31

6

x

632

31

31

xx ? 9 2/3 = 6





1 f(x) =

2 f(x) =

3 g(x) =

4 h(x) =


1 f (x) =

2 f (x) =

3 g(x) =

4 h(x) =

43

12x

52

5x

31

3x 32

x

41

2x 43

21 x

41

9 x

53

x




When a function is given in the form of an equation, the derivative

is written in the form dx

dy

1 f(x) = 3x2

f (x) = 6x y = 3x2

dxdy

= f (x) dxdy

2 f(x) = 4x -3

f (x) = -12x -4 y = 4x -3

= 6x = -12x -4 dxdy

Examples





End of Section 5




ff ((xx))dydydxdx


6 Sums of functions (simple cases)




More examples

f(x) = 5x2 - 3x + 1

f (x) =

g(x) h(x) k(x)

- 3 + 010x

= 10x - 3

g(x) = (x + 3)(x - 2)

= x2 + x - 6

g (x) = 2x + 1

1 2





If f(x) = g(x) + h(x) + k(x) +……... , then f (x) = g (x) + h (x) +k (x)

+..





1 f(x) = x2 + 7x - 3

2 f(x) = 3x2 - 4x + 10

3 g(x) = x(x2 + x - 5)

4 h(x) = 4x3 - 10x2

5 f(x) = x5(7 - 5x2)

6 f(x) = x3 + x2 + x + 1


1 f (x) = 2x + 7

2 f (x) = 6x - 4

3 g(x) = 3x2 + 2x - 5

4 h(x) = 12x2 - 20x

5 f (x) = 35x4 - 35x6

6 f (x) = 3x2 + 2x + 1

7 f(x) = ¼ x4 +½ x2 7 f (x) = x3 + x




Now do Section A2 on page 33 of the Basic Skills booklet

End of Section 6




ff ((xx))dydydxdx


7 Sums of functions (negative indices)




More examples

f(x) =

f (x) =

=

2

2x x

5- - 3x

= 2x -2 - 5x -1 - 3x

-4x -3 + 5x -2 - 3

3

4x 2

5x

- 3- +

y = 2

1x

+ -x1

1

= x -2 - x -1 + 1

dxdy

= -2x -3 + x -2 + 0

= - 3

2x + 2

1x

1 2





1 f(x) =

2 y =

3 y =

4 g(x) =


1 f (x) =

2

3

4 g(x) =

2

2x

- 2x

x3

+ x2 - x

3

2x

- 2

1x

+ 3x2

x1

4

3x

+ 5

2x

-

3

4x

- - 2

2

3x

= - + 2x - 1dxdy

dxdy

= 4

6x

- + 3

2x

+ 6x

- 5

12x

- 6

10x

+ 2

1x





End of Section 7




ff ((xx))dydydxdx


8 Sums of functions (algebraic fractions)




More examples

f(x) =

1

xxx 352

=xx2

+xx5

-x3

= x

f (x) = 1 3x -2

= 1 + 2

3x

2

y = 2

2 23x

xx

= + -2

2

xx

2

3xx

2

2x

= 1

dxdy

= 0 - 3x -2 + 4x -3

= 2

3x

- + 3

4x

+ 0

+ 5 - 3x -1

+

+ 3x -1 - 2x -2





1 f(x) =

2 y =

3 g(x) =

4 y =


1 f (x) = 2x - 2

2

3 g(x) = 1 +

4

xxx 23

xxx 52

dxdy

= 1

xxx 132

2

1x

2

2 52x

xx dxdy

= 2

1x

- + 3

10x




Now do Section A3 on page 33 of the Basic Skills booklet

End of Section PC(a)



PC(b) Determine the gradient of a tangent to a curve by differentiation

dydydxdx

ff ((xx))PC(b) -

Gradientof a tangent





The gradient of a tangent to thecurve y = f (x) is f (x) or

dxdy




Find the gradient of the tangentto the curve y = x2 - 5x at each of the points A and B.

Gradient of tangent = dxdy

= 2x

When x = 3 , dxdy

= 2(3) = 1

When x = -1, dxdy

= 2(-1) = -7

y

x

A (3, -6)

B (-1, 6)

y = x2 - 5x

So gradient of tangent at A is 1and gradient of tangent at B is -7

m = 1

m = -7

A

B

Example 1

- 5

- 5

- 5




Example 1 (continued) y

x

A (3, -6)

B (-1, 6)

y = x2 - 5x

m = 1

m = -7m = 1

Point on line is (3, -6)

Equation of tan is y - b = m(x - a)

y

y

y =

Find the equation of the tangentto the curve y = x2 - 5x at each of the points A and B.

Tangent at A

- (-6)= 1 (x - 3)

+ 6 = x - 3

x - 9

y = x - 9




Find the equation of the tangentto the curve y = x2 - 5x at each of the points A and B.

y

x

A (3, -6)

B (-1, 6)

y = x2 - 5x

m = 1

m = -7Tangent at B m = -7

Point on line is (-1, 6)

Example 1 (continued)


y -

y - 6 =

y =

NB 3 negatives

6 -7(x - (-1))=

-7x - 7

-7x - 1

y = x - 9

y = -7x - 1




Find the gradient of the tangentto the curve y = x3 - 5x + 3at each of the points P and Q.

Gradient of tangent = dxdy

= 3x2 - 5

When x = 1 , dxdy

= 3(1)2 - 5 = -2

When x = -2, dxdy

= 3(-2)2 - 5 = 7

y

xP (1, -1)

Q (-2, 5)

So gradient of tangent at P is -2and gradient of tangent at Q is 7

m = -2

m = 7

P

Q

Example 2

y = x3 - 5x + 3





xm = -2

Point on line is (1, -1)


y -

y

y =

y = x3 - 5x + 3

P (1, -1)

Q (-2, 5)

m = -2

m = 7

Tangent at P

Find the equation of the tangentto the curve y = x3 - 5x + 3at each of the points P and Q.

(-1) = -2 (x - 1)

+ 1 = -2x + 2

-2x + 1

y = -2x + 1





xTangent at Q m = 7

Point on line is (-2, 5)


y -

y - 5 =

y =

Find the equation of the tangentto the curve y = x3 - 5x + 3at each of the points P and Q.

y = x3 - 5x + 3

P (1, -1)

Q (-2, 5)

m = -2

m = 7

5 = 7 (x - (-2))

7x + 14

7x + 19

y = -2x + 1

y = 7x + 19




1 Find the gradient of the tangent to the curve y = x2 + 5 at each of the points A and B.

Find the equation of each tangent.

y

x

A (2, 9)

B(-4, 21)

Tangent at Am = 4Equation is y = 4x + 1

Tangent at Bm = -8Equation is y = -8x - 11

Answers

y = x2 + 5

dxdy

= 2x




2 Find the gradient of the tangent to the curve y = (3 + x)(3 - x) at each of the points E and F.


y

x

F

Tangent at Em = -4Equation is y = -4x + 13

Tangent at F (-3, 0)m = 6Equation is y = 6x + 18

Answers

y = (3 + x)(3 - x)

E (2, 5)

dxdy

= -2x




3 Find the gradient of the tangent to the curve y = x4 - 5x2 + 4 at each of the points A , B and C.


y

x

A (0, 4)

B(-2, 0)

Tangent at Am = 0

Equation is y = 4

Tangent at Bm = -12

Equation is y = -12x - 24

Answers

C(1, 0)

Tangent at Cm = -6

Equation is y = -6x + 6

y = x4 - 5x2 + 4

dxdy

= 4x3 - 10x




Now do Sections B1and B2 on page 37 of the Basic Skills booklet

End of Section PC(b)



PC(c) Determine the coordinates of the stationary points on a curve…….

dydydxdx

ff ((xx))PC(c) -

Stationary points





Stationary points occur when f (x) = 0 or = 0

dxdy




To determine the nature of a stationary point, find out the gradient of the tangent before and after the stationary point.

before after

y

x

dxdy

dxdy

dxdy

+ ve - ve

= 0 x before after

dxdy

+ 0 -

slope

This is a maximum turning point





before after

y

x

dxdy

dxdy

dxdy

- ve + ve

= 0

x before after

dxdy

- 0 +

slope

This is a minimumturning point





x before after

dxdy

- 0

slope

This is a minimumturning point

x before after

dxdy

+ 0 -

slope

This is a maximum turning point

+



y = x2 - 6x + 5

dxdy

= 2x

For stationary values,dxdy

= 0

ie 2x - 6 = 0 x = 3

When x = 3,

(3)2

= 9

= -4

So a stationary point

occurs at (3, -4).

y =

Find the stationary point on the curve with equation y = x2 - 6x + 5.

Example 1

Using differentiation determine its nature.


-6(3) + 5-18 + 5

- 6



Find the stationary point on the curve with equation y = x2 - 6x + 5.

Example 1 (continued)

A stationary point occurs at (3, -4).

x 3 -

dxdy

- 0

slope

(3, -4) is a minimum turning point.

Using differentiation determine its nature.

3 3 +

= 2x - 6

+

dxdy




dxdy

= x2

For stationary values, dxdy

= 0

ie x2 - 2x - 3 = 0 ( )( ) = 0

When x = 3,

1/3(3)3

So stationary points occur at (3, 1) and(-1, )

y =

Find the stationary points on the curve with equation y = 1/3x3 - x2 - 3x + 10.

Example 2

Using differentiation determine their nature.

- 2x - 3

x =

When x = -1,y =1/3(-1)3 3

211

3211


1

x =x - 3 x + 1

-32 -3(3)+10 =

-(-1)2-3(-1)+10=

3, -1




Example 2


x 3 -

dxdy

- 0

slope

(3, 1) is a minimum turning point.

3 3 +

+A stationary point occurs at (3, 1).

dxdy

x2 - 2x - 3=





Example 2


x -1 -

dxdy

+ 0

slope

(-1, ) is a maximumturning point.

-1 -1 +

-A stationary point occurs at (-1, ).

dxdy

x2 - 2x - 3=

3211


3211



The graph of the curve y = 1/3x3 - x2 - 3x + 10

looks something like this:

Example 2

y

x

2

4

6

0 1 2 3 4

8

10

12

-4 -3 -2 -1 5

(3, 1)

(-1, )3211

Maximum

turning point.

Minimum turning point.

-5

y = 1/3x3 - x2 - 3x + 10





1 Find the stationary point on

the curve y = x2 + 3 .

Using differentiationdetermine its nature.

Solution

dxdy

= 2x


= 0

ie 2x = 0

x = 0

When x = 0,

(0)2 + 3y =

= 0 + 3


occurs at (0, 3).

= 3





the curve y = x2 + 3 .


Solution (continued)

dxdy

= 2x

A stationary point occurs at (0, 3).

x 0 -

dxdy

- 0

slope

0 0 +

+

(0, 3) is a minimum turning point.





the curve y = 16 - x2 .


Solution

dxdy

= -2x


= 0

ie -2x = 0

x = 0

When x = 0,

16 - (0)2 y =

= 16 - 0


occurs at (0, 16).

= 16





dxdy

= -2x

A stationary point occurs at (0, 16).

x 0 -

dxdy

+ 0

slope

0 0 +

-

(0, 16) is a maximumturning point.


the curve y = 16 - x2.






the curve y = (x + 5)(x - 3) .


Solution

dxdy

= 2x

For stationary values,dxdy

= 0

ie 2x + 2 = 0

x = -1

When x = -1,

(-1)2y =

= 1


occurs at (-1, -16).

= -16

y = x2 + 2x - 15

+ 2

- 15+2(-1)

- 2 - 15





the curve y = (x + 5)( x - 3) .



dxdy

= 2x + 2

A stationary point occurs at (-1, -16).

x -1 -

dxdy

- 0

slope

-1 -1 +

+

(-1, -16) is a minimum turning point.




4 Find the stationary points on

the curve y = 1/3 x3 - x2 - 15x.

Using differentiationdetermine their nature.

Solution

dxdy

For stationary values, = 0dxdy

= x2 - 2x - 15

ie ( )( ) = 0x - 5 x + 3

x = x =

When x = 5,y = 1/3(5)3 - (5)2 - 15(5) = 3

158

When x = -3,

y = 1/3(-3)3 -(-3)2 -15(-3) = 27

So stationary points occur at (-3, 27) and(5, )3

1585, -3





the curve y = 1/3 x3 - x2 - 15x.



dxdy

=

A stationary point occurs at (-3, 27).

x2 - 2x - 15

x -3 -

dxdy

+ 0

slope

-3 -3 +

-

(-3, 27) is a maximum turning point.





the curve y = 1/3 x3 - x2 - 15x.



dxdy

=

A stationary point occurs at (5, ).

x2 - 2x - 15

x 5 -

dxdy

slope

5 5 +

(-3, ) is a minimum turning point.

3158

3158

- 0 +




Now do Sections C1and C2 on page 43 of the Basic Skills booklet

End of Section PC(c)

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Basic Skills in Higher Mathematics