derivatives

DERIVATIVES

Definition:

If f(x) is a function, then the derivative of f(x) is:

Example:

Find the derivative of f(x) = x² + 1

Answer:

Notations of derivative: .

THE FORMULAS OF DERIVATIVE

1. F(x) = c, c R then .

Example:

f(x) = 5 then f’(x) = 0.

2. F(x) = ax + b, then .

Example:

F(x) = 3x then f’(x) = 3.

3. F(x) = axⁿ then f’(x) = n.ax^n-1.

Example:

Find the derivatives of the following functions:

a. F(x) = x² b. f(x) = x³ c. f(x) = 4x⁶

Answer:

a. F(x) = x², then f’(x) = 2x.

b. f(x) = x³, then f’(x) = 3x².

c. f(x) = 4x⁶, then f’(x) = 24x⁵.

4. F(x) = u(x) v(x) then f’(x) = u’(x) v’(x).

Example:

Find the derivatives of the following functions:

a. F(x) = 3x² + 6x + 2 b. f(x) = x³ – 5x² + c. f(x) = (3x+1)² d. f(x) =

Answer:

a. F(x) = 3x² + 6x + 2 c. f(x) = (3x+1)² = 9x² + 6x +1

F’(x) = 6x + 6. f’(x) = 18x + 6.

b. f(x) = x³ – 5x² + d. f(x) =

= x³ – 5x² + 8x^-1 = =

f’(x) = 3x² – 10x – 8x^-2 f’(x) =

= 3x² – 10x - . = = .

5. y = u.v then y’ = u’.v + v’.u. (u and v is a function)

Example:

Find the derivatives of:

a. f(x) = (x² – 1)(3x + 2) b. f(x) = (x⁴ + 5)(x² + 3)

Answer:

a. f(x) = (x² – 1)(3x + 2) b. f(x) = (x⁴ + 5)(x² + 3)

u = x² – 1 → u’ = 2x u = x⁴ + 5 → u’ = 4x³

v = 3x + 2 → v’ = 3 v = x² + 3 → v’ = 2x

f’(x) = u’.v + v’.u f’(x) = u’.v + v’.u

= 2x.( 3x + 2) + 3.( x² – 1) = 4x³.( x² + 3) + 2x.(x⁴ + 5)

= 6x² + 4x + 3x² – 3. = 4x⁵ + 12x³ + 2x⁵ + 10x

= 6x⁵ + 12x³ + 10x.

6. then .

Example:

Find the derivative of .

Answer:

.

7. F(x) = {u(x)}ⁿ then f’(x) = n.{u(x)}^n-1.u’(x)

Example:

Find the derivatives of :

a. F(x) = (3x + 1)⁵ b. f(x) =

Answer:

a. F(x) = (3x + 1)⁵ b. f(x) = =

F’(x) = 5.(3x + 1)⁴.3 f’(x) =

= 15(3x + 1)⁴. = (x+2).

TANGENT EQUATION

The tangent equation through point (x₁,y₁) with the slope (gradient) m:

y – y₁ = m(x – x₁)

where m =f’(a) = (the derivative of y for x = a)

Examples:

1. Find the tangent equation on curve y = x³ + 2x through point (1,3)

2. Find the tangent equation on curve y = x² – x + 3 through absis point 2

3. Find the tangent equation on curve y = x² + 4 with the slope 8.

Answer:

1. y = x³ + 2x 3. y = x² + 4 m = 8

y’ = 3x² + 2 y’ = 2x ↔ 8 = 2x

(1,3) then x = 1 → m = y’ = 3.(1)² + 2 = 5. x = 4

Tangent equation through (1,3) with m = 5 x = 4 → y = (4)² + 4

y – y₁ = m(x – x₁) = 20.

y – 3 = 5(x – 1) (4,20)

y – 3 = 5x – 5 T.e. through (4,20) with m = 8

y = 5x – 2. y – y₁ = m(x – x₁)

So, the tangent equation is y = 5x – 2. y – 20 = 8(x – 4)

2. y = x² – x + 3 y – 20 = 8x – 32

y’ = 2x – 1 y = 8x – 12.

x = 2 → m = y’ = 2.(2) – 1 = 3. So, the tang. Eq. is y = 8x – 12.

X = 2 → y = (2)² – 2 + 3 = 5. (2,5)

Tangent equation through (2,5) with m = 3

y – y₁ = m(x – x₁)

y – 5 = 3(x – 2)

y – 5 = 3x – 6

y = 3x – 1.

So, the tangent equation is y = 3x – 1.

Notes:

Parallel lines → m₁ = m₂

Perpendicular → m₁.m₂ = -1

Examples:

1. Find the tangent equation of touching curve y = x² – x + 1 which parallel to y – 5x + 2 = 0.

2. Find the tangent equation of touching curve y = x² + 2x + 4 which perpendicular to x – 4y + 3 = 0.

Answer:

1. line y – 5x + 2 = 0 m₁ = 5 x = 3 y = (3)² – 3 + 1 = 7

parallel m₁=m₂=5 tangent point (3,7)

y = x² – x + 1 tangent equation through (3,7) with m = 5

y’ = 2x – 1 y – y₁ = m(x – x₁)

5 = 2x – 1 (m₂=5) y – 7 = 5(x – 3)

2x = 6 y – 7 = 5x – 15

x = 3 y = 5x – 8

2. line x – 4y + 3 = 0 m₁ = x = -3 y = (-3)² + 2(-3) + 4 = 7

perpendicular m₁.m₂ = -1 tangent point (-3,7)

. m₂ = -1 tangent equation through (-3,7) with m = -4

m₂ = -4 y – y₁ = m(x – x₁)

curve y = x² + 2x + 4 y – 7 = -4(x + 3)

y’ = 2x + 2 y – 7 = -4x -12

-4 = 2x + 2 (m₂ = -4) y = -4x -5

2x = -6 So, the tangent equation is y = -4x -5

x = -3

THE FUNCTION INCREASES AND THE FUNCTION DECREASES

F(x) increases if f’(x) > 0.

F(x) decreases if f’(x) <>

Examples:

1. Find the interval when f(x) = x³ + 3x² + 5 decreases!

2. Find the interval when f(x) = 5 + 15x + 9x² + x³ increases!

Answer:

1. F’(x) = 3x² + 6x <>

x² + 2x <>

x(x + 2) <>

x = 0 / x = -2

+++ - - - +++

-2 0

-2 < style=""> So, the decreasing interval is -2 <>

2. F’(x) = 0 + 15 + 18x + 3x² > 0

3x² + 18x + 15 > 0

x² + 6x + 5 > 0

(x + 5)(x + 1) > 0

x = -5 / x = -1

+++ - - - +++

-5 -1

x < -5 x > -1

So, the decreasing interval x < -5 or x > -1

STATIONER POINT

F(x) stationer if f’(x) = 0.

Types of stationer points:

1.

[a,f’(a)]

Maximum point Point of inflexion (titik belok)

f’(a) is called maximum value.

a

[a,f’(a)] is called maximum point.

2. Minimum point

f’’(x) <>

f’(a) is called minimum value.

[a,f’(a)] is called minimum point.

[a,f’’(a)] is called inflexion point.