derivative of a function
DESCRIPTION
Mathematics Departament ISV International School of Valencia BRITISH SCHOOL EL PLANTÍO Differentation Review Presentation Y12 Curriculum Maths Teachers: José Ramón Fierro, Head. Ignacio Muñoz Motilla. Derivative of f(x) = x 2 in x=3:. Derivative of a function. f(x) = x 2. - PowerPoint PPT PresentationTRANSCRIPT
Mathematics DepartamentISV International School of Valencia
BRITISH SCHOOL EL PLANTÍO
Differentation Review PresentationY12 Curriculum
Maths Teachers:José Ramón Fierro, Head.
Ignacio Muñoz Motilla.
• Notice that the expression f ´(x) is itself a function and for this reason we also refer to the derivative as the gradient function of y = f(x).
Derivative of a function
f´(3) = h 0lim
f(3 +h)-f(3)h =
h 0lim
(3 + h)2-32
h = h 0lim
h(h + 6) h = 6
• Derivative of f(x) = x2 in x=2:
• To get the derivative in x=2:
f´(x) = h0lim
f(x+ h)-f(x) h =
h 0lim
(x+ h)2-x2
h = h 0lim
h (h +2x) h = 2x
• Derivative of f(x) = x2 in x=3:
f´(2) = h 0lim
f(2 +h)-f(2)h =
h 0lim
(2 + h)2-22
h = h 0lim
h(h + 4) h = 4
X
Y
f(x) = x2
X
Y
f ´(x) = 2x
X
Y
f (x)
f ' (x)
Some examples of derivative functions
Derivative of f(x) = k is f ' (x) = 0
Derivative of f(x) = x is f ' (x) = 1X
Y f (x)
f ' (x)
Cannot get the derivative
Continuous function 0lim ( ) ( )h
f x h f x
• If a function has a derivative in one point “P”, if exists the straight line tangent (not vertical) in this point to the graph, that means that the graph in this point is continuous.
Derivative function (Differentation) and continuity
0
( ) ( )limh
f x h f xC
h
0h
0lim ( ) ( ) 0h
f x h f x
• Absolute value function f(x)=|x| is continuous in , but has no tangent line in x=0, thta means we cannot get the derivative.
0lim ( ) 0 (0)x
f x f
0
0
(0 ) 0(0 ) lim 1
(0 ) 0(0 ) lim 1
h
h
hf
hh
fh
Rules for differentiation
y = a . f(x) y ' = a . f '(x)
y = f(x) g(x) y ' = f '(x) g '(x)
y = f(x) . g(x) y ' = f '(x) . g(x) + g '(x) . f(x)
y = f(x)
g(x)y ' =
f '(x) . g(x) – f(x) . g '(x)
g2(x)
y = f(x) (f–1(x))' = 1
f '(y)/ y = f(x)
The Chain Rule
y = f [g(x)] y ' = f ' [g(x)] .g'(x)
y ' = (sen ' t) (t )' = cos t . 2 = cos 2x . 2
y ' = (2t)' (t)' = 2. cos x
2x = t
y = sen 2x
y = 2 sen x
sen x = t
DERIVATIVE OF RECIPROCAL CIRCULAR FUNCTIONS (I)
Let’s get the derivative of function 1( )f x
11. ( )( )f f x x
2. With the chain rule
(f o g)’ (x) = f ’ (g(x)) g’(x)} 1 1( ( )) ( ) ( ) 1f f x f x
11
1( ) ( )
( ( ))f x
f f x
DERIVATIVE OF RECIPROCAL CIRCULAR FUNCTIONS (II)
X
Y
P(y, x)
P ' (x, y)
f
f–1
• •
a
90 – a
• f ' (y) = tg a
• (f –1(x))' = tg (90 – a) = 1 / tg =a1 / f '(y) con f –1(x) = y
DERIVATIVE OF LOGARITHMIC FUNCTIONS
Let’s calculate the derivative of :log ( )a x
} log
1( )
ln a xg xa a
1
ln a x
log ( )a x
1. ( )( ) ( )( ) .f g x g f x x
Using the reciprocal of a Function rule
Sean ( ) y ( ) log ( ).xaf x a g x x
Then the derivitaive of
Will be;
X
Y
Monotony: Growth and decrease in a rangeRegarding Average and Instantaneous rate of Change (ARC and IRC)
[a
]b
x
f(x)
x+h
f(x+h)h
Increasing function in [a, b]
f(x) < f(x+h), (x, x+h) y h >0
ARC (x, h) > 0 (x, x+h) y h >0
X
Y
[a
]b
x
hf(x)
Decreasing function in [a, b]
–ARC(x,h)
f(x) < f(x+h), (x, x+h) y h >0
ARC (x, h) < 0 (x, x+h) y h >0
f(x+h)
x+h
ARC(x,h)
STATIONARY POINTS
X
Y
f ' < 0 f ' > 0 f ' < 0a
b
f ' (a) = 0f " (a) > 0
f " (b) < 0f ' (b) = 0
Local minimum of coordenates (a, f(a))
Local maximum ofcoordenates (b, f(b))
So far we have discussed the conditions for a function to be increasing ( f '(x) > 0) and for a function to be decreasing ( f '(x) < 0). What happens at the point where a function changes from an increasing state (( f '(x) > 0) ) to ( f '(x) = 0) and then to a decreasing state (( f '(x) < 0) ) or vice–versa?Points where this happens are known as stationary points. At the point where the function is in a state where it is neither increasing nor decreasing, we have that f '(x) = 0 . There are times when we can call these stationary points stationary points, but on such occassions, we prefer the terms local maximum and local minimum points.
Derivative of Sinus Function
Let’s calculate the derivative of sen( )x
0
sen( ) sen( )(sen( )) = lim =
h
x h xx
h
0
0
Como
lim cos cos( )2
2 lim 1
2
h
h
hx x
hsen
h
The derivative of will be cos( )x
sen( )x
Using the derivative definition
0
2 cos sen2 2
lim h
h hx
h
Derivative of the Tangent Function
Let’s calculate the derivative of tg( )x
2
sen( ) cos( ) sen( ) cos( )(tg( )) = =
cos ( )
x x x xx
x
The derivative is
21 tg ( )x
tg( )x
2 2
2
cos sen
cos ( )
x x
x
Using the formulasen( )
tg( )= cos( )
xx
x
Derivative of the arc sinus function
Let’s calculate the derivative of arcsen( )x
} 1( )
cos(arcsen( ))g x
x
The derivative is
2
1
1 xAs it is,
2 2cos(arcsen ) 1 sen (arcsen ) 1x x x
Knowing ( ) sen( ) y ( ) arcsen( ).f x x g x x
1. ( )( ) ( )( ) .f g x g f x x
Using the reciprocal of a Function rule
Derivative of the arc tangent function
Let’s calculate the derivative of arctg( )x
} 2
1( )
1 tg (arctg( ))g x
x
The derivative will be
2
1
1 x
As, it is
( )tg arctg x x
1. ( )( ) ( )( ) .f g x g f x x
Knowing that ( ) tg( ) y ( ) arctg( ).f x x g x x
11
2. Derivada función recíproca
1 ( ) ( ) .
( ( ))f x
f f x
Using the reciprocal of a Function rule
y = f n(x) y '= n . f n–1(x) . f '(x)
y = loga[f(x)] y ' = f '(x)
f(x) · loga e
y = af(x) y ' = af(x) · f '(x) · ln a
y = sen f(x)
More rules
Function Its derivative function
y ' = cos f(x) . f '(x)
y = cos f(x) y ' = – sen f(x) . f '(x)
y = tg f(x) y ' = f '(x)
Cos2 f(x)
y = arcsen f(x)
y = arctg f(x)
y = arccos f(x)
y = arcctg f(x)
More rules (II)
Function Its derivative
y ' = 1 + f2(x)
– f '(x)
y ' = 1 – f2(x)
f '(x)
y ' = 1 – f2(x)
-f '(x)
y ' = 1 + f2(x)
f '(x)
Curvature: Convexity and Concavity
X
Y
[a
]b
X
Y
[a
]b
X
Y
[a
]b
X
Y
[a
]b
Average Rate of Change positive and increasing: Convex function Average Rate of Change negative and decreasing: Convex function
Average Rate of Change positive and increasing: Concave function Average Rate of Change negative and decreasing: Concave function
X
Y
[a
]b
Relations between the derivative function and curvature
The gradients of the function increase f ' is increasing f " > 0 convex function
X
Y
[a
]b
a1
a2
x1 x2
tg a1<tg a2 f '(x1) < f '(x2)
x1 x2
a1
a2
X
Y
[a
]b
a1
a2
X
Y
[a
]bx1 x2
a1
a2
x1 x2
tg a1>tg a2 f '(x1) > f '(x2)
The gradients of the function decrease f ' is decreasing f " < 0 Concave function
Relations between the derivative function and curvature
Stationary Point of Inflection
X
Y
P(a, f(a))
f" < 0
f" > 0
f"(a) = 0
Summary regarding plotting a Graph of a Function
1. Study domain and continuity.
3. Intersection points with both axis
4. Get possible asymptotes.
5. Monotony. Study first derivative
6. Curvature. Get second derivative {
{Vertical: Points that are not in the domain.
Horizontals or obliquess: Getting limits in the infinity.
2. Check simetry and periodicity.
{X-axis
Y-axis:
f (x) = 0
f (0)
{Posible stationary:
Growth:
Decreasing:
f ‘ (x) = 0
f ‘ (x) > 0f ‘ (x) < 0
Posible Inflection Points:
Convex:
Concave:
f “ (x) = 0f “ (x) > 0
f “ (x) < 0
Plotting polynomial functions (I)
Let’s plot the following function: 3( ) 4f x x x R is its domain, it’s continuousand has no asymptotes
1. Interception points with both axis
2. Simetry
3. Limits in the infinity
Y-axis: 3(0) 0 4 0 0f (0,0)
X-axis : 3 4 0; =0, 2x x x {( 2,0)
(0,0)
(2,0)
3 3( ) ( ) 4( ) ( 4 ) ( )f x x x x x f x
3
3
lim 4
lim 4
x
x
x x
x x
ODD
Sketching and Plotting the function 3( ) 4f x x x 4. Monotony
2( ) 3 4f x x
23 4 0x
if 2 3( ) 0
3x f x
2 2 3
33x
if 2 3 2 3( ) 0
3 3x f x
if 2 3( ) 0
3x f x
Plotting polynomial functions (II)
Sketching and plotting 3( ) 4f x x x 5. Curvature
( ) 6f x x
6 0x
if 0 ( ) 0x f x
0x
if 0 ( ) 0x f x
Plotting polynomial functions (III)
Plotting Rational Functions (I)
Let’s plot the following function
4( )
4(1 )
xf x
x
1. Domain and continuity
2. Interception points with axis
3. Simetry
{1}R
( ) ( )f x f x
It has not
4(1 ) 0; 1x x
Y axis:0 4
(0) 14(1 0)
f
(0, 1)
X axis: 4 0; =4x x (4,0)
6. Curvature
3
3( )
2( 1)f x
x
if 1 ( ) 0x f x
if 1 ( ) 0x f x
Plotting Rational Functions (II)
The is not any stationary point of inflection
Plotting and sketching the function
4( )
4(1 )
xf x
x