DEFINITION OF DERIVATIVE
DEFINITION OF DERIVATIVE • Linear function -gradient of a linear function: (x1,y1) a (x0,y0)
y1 y 0 a m x1 x 0 b
b
- Derivative-gradient of the tangent to a curve 2
DEFINITION OF DERIVATIVE Secant line Q(x+x,y+y) y P(x,y) x
N
• The average rate of change of y with respect to x on the interval [x, x+x] is given by •
change in y Average rate of change , rave = change in x
δy = δx
f(x δx) f(x) = δx 3
DEFINITION OF DERIVATIVE f(x δx) f(x) δx
From the graph of the function, we see that which is both the average rate of change and slope of the line PQ.
Secant Line: The slope of the secant line PQ: Slope of PQ =
QN δy f(x δx) f(x) PN δx δx
Tangent line : The slope of the tangent line at P -If y = f(x), the instantaneous rate of change of y with respect to x ,is given by the derivative of f: rinst = f’(x)= lim f ( x x ) f ( x ) x
x 0
f(x) Q x x, y y
y y
-As ∂x
0, Q
mPQ
P, PQ
P tan
f (x x ) f (x ) x as x 0
y
- Therefore, The slope of the tangent line at P
P x, y
tangent lim
x
0 x
x x
-This slope producing function, called derivative of f with respect to x and its notation- f’(x).
x 0
f x x f x x
Tangent line : The slope of the tangent line at P Derivative of a function: • The derivative of a function f is the function f ’(x) defined by
f' (x)
dy dx
δy f(x δx) f(x) lim δx 0 δx δx 0 δx
lim
provided that the limit exists. If f’(x) exists, then we say that f is differentiable at x • Or using h instead of x, the formula is in the form
f ( x h) f ( x ) f ' ( x ) lim h 0 h This process is called differentiation from first principles.
Interpretations of the derivatives: • The derivative of a function f is a new function f’. The derivatives have various applications and interpretations, including the following: 1. Slope of the tangent line. For each x in the domain of f’, f’(x) is the slope of tangent line to the graph of f at the point (x, f(x)). 2.
Instantaneous rate of change. For each x in the domain of f’, f’(x) is the instantaneous rate of change of y =f(x) with respect to x.
3.
Velocity. If f(x) is the position of the moving object at time x , then v=f’(x) is the velocity of the object at that time.
The steps to find the derivative from the first principles : • The steps process for finding the derivative of a function f using differentiation from first principles: Step 1 : Step 2 :
Find
f(x) and f(x+h)
Find
f ( x h) f ( x ) lim h0 h
Example 19 Using the first principle find the derivatives with respect to x : a) y = 5x +3
b) f(x) = 4x2 – 2
1 c) y = 3 x 1
d) f(x) = x 1
TECHNIQUES OF DIFFERENTIATION
Rules of differentiation, higher order derivatives 1. 2.
d (a) 0 dx
where a is a constant
d n n1 ( x ) nx Example 20 dx Differentiate the following functions with respect to x: 3.
a) y = e) y =
b) f(x) = x10
c) f(x) = 5x6
d f ) g) y = dh) y= k f (x) k f (x) dx dx
d) y = 4x 5
3
x
3
y4 x
2
x3
3 2x 4
a) Differentiation of sums and differences 1. Sum of function : 2. Difference of function : Example 21
d f ( x ) g( x ) d f ( x ) d g( x ) dx dx dx d f ( x ) g( x ) d f ( x ) d g( x ) dx dx dx
Differentiate the following with respect to x: a) f(x) = x3+4x2-9x-13 b) c) f(x)=(x – 5 ) 2 d) f(x) = (2x-1)(x+3)
2 3
f ( x ) 5x 3x e) f)
x3 1 f (x) x2
1 4 x
f x x ( x 2)
2 1 3 x
b) Differentiation of composite functions (Chain rule)
3. Chain rule or function of a function rule
Example 22 Find the differentiation with respect to x. (a) (2x-3)7 (b) (x2+ 1)4 (c)
If y = f(u) and u = g(x), hence y = f(u) = f(g(x)). If f(x) and g(x) is differentiable, then dy du f (u) dan g( x ) du dx Thus dy dy . du dx du dx Power rule – another version of chain rule (d)
d (e) f (x ) dx
n n f ( x ) n 1 d f ( x )
1 4x 5
dx
2
3x 1
(3 x
2
1 5 1)
c) Differentiation of Product
4. Product rule Example 23 Differentiate the following with respect to x: a) (4x2 - 1)(7x3 + x) b) (2 x + 3)( x – 1)4 c)
d) (2x+3)5(x3+3)10
x x2
If y = uv, where u and v are functions in terms of x, then dy dv du u v dx dx dx
d) Differentiation of Quotients
5. Quotient rule If y =
u where u and v are functions in terms of x, then v du dv
Example 24 Differentiate the following with respect to x : a) b) Example 25 Differentiate evaluate
by using chain rule with u = when
v u dy dx dx . Hence dx v2
.
x( x 1) x3 1
2x 2 1 x2 1
1 x 1 x dy dx
1 x 1 x x
1 2
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS d (sin x ) cos x dx
d (cos x ) sin x dx d (tan x ) sec 2 x dx Example 26 Differentiate the following functions with respect to x: (a)
2 sinx – 3 cosx
5 (b) 4 tan x x
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS d du (sin (ax b)) a cos (ax b) cos u dx dx d du (cos (ax b)) a sin (ax b) sin u dx dx d (tan (ax b)) a sec 2 (ax b) dx
Example 27 Differentiate the following with respect to x : a) sin 5x
b)
1 cos ( 1 x ) 3
d) 4 cos 3x
e)
y = sin(3x+b)
c)
1 35 tan x 7
f)
tan(x2+)
DERIVATIVES OF EXPONENTIAL FUNCTIONS d x (a ) a x ln a dx d f(x) 2. (a ) a f(x) ln a.f'(x) dx d x 3. (e ) e x dx d f(x) 4. (e ) e f(x).f ' (x) dx
1.
Example 28 Differentiate each of the following with respect to x : 1 x 3 x 2 3 x -4x 3x+1 sin x a)e b) e c) e d) e e) 2 f) 3cos x Example 29 dy Find for dx
a) y = xe
x
b) y =
3x 2 e 2x
DERIVATIVES OF LOGARITHM FUNCTIONS d 1 1. (ln x ) dx x 2. Note:
d f ' (x) (ln f (x )) dx f (x)
d 1 (ln x ) dx x
The following laws of logarithm is used to simplify any logarithmic functions before it is differentiated. This is to make it easier to differentiate. ln (mn) = ln m + ln n ln (m/n) = ln m – ln n ln (m)p = p ln m
Example 30 Differentiate each of the following with respect to x : a) ln (3x) d) ln (2x+1)
3
5x 2 ln g) 2 x 3x 2
b) ln (x2 - 6 )
c) ln (sin x)
e) ln (5x-1)(3x+8)
x 2 1 f) ln 2 x 1
h) 2x3ln(3x-2)
IMPLICIT DIFFERENTIATION
Definition: 1.
Explicit function : y = f(x) variable y appears alone at one side of the equations.
2.
Implicit function : F(x,y) = c where c is a constant variable y cannot be stated as a subject
dz dz 2y . But what is If z = y , thus ? dy dx 2
dz dz dy dy . 2y By the function of a function rule, dx dy dx dx
IMPLICIT DIFFERENTIATION Some implicit differentiation of y with respect to x : 1.
dy dy d (y) 1 dx dx dx
2.
d d 2 1 dy ( y) (y ) dx dx 2 y dx
3.
d d d dy ( xy) y ( x ) x ( y) y x dx dx dx dx
1
IMPLICIT DIFFERENTIATION Example 31 Find : a) x2 + y2= 25
b) x 3+y 3 +3xy =9
c) 4xy3-x2y+x3-5x+6=0
d) (x+y)5-7x2=0