The Derivatives

The average rate of change of y w.r.t x on the interval [x, x+δx ] is

given by

Average rate of change f(x)

f x  δx 

change in y  change in x

f(x) Qx  δx, y  δy 

y  x δy

f(x)

P (x,y) δx

x

f(x  δx)  f(x)  x  δx  x

N x  δx

x

f(x  δx)  f(x)  δx

For example, the average velocity is the average rate of change of the displacement, s w.r.t t in some interval of time.

From the graph of the function,

f(x  δx)  f(x) δx which is both the average rate of change and slope of the line PQ.

The line PQ is called the secant line.

Suppose that we wish to find the gradient of a curve y = f( x ) at a

point P(x, y ) on the curve. f(x) f x  δx 

Gradient of PQ

f(x) Qx  δx, y  δy 

QN  PN

change in y change in x

δy

f(x)

P (x,y) δx

x

f(x  δx)  f(x)  x  δx  x

N x  δx

x

f(x  δx)  f(x)  δx

QN δy f(x  δx)  f(x)   Slope of PQ = PN δx δx

Secant Line: The slope of the secant line PQ:

If we move Q nearer to P , say to points Q1, Q2 ,…

f(x) f x  δx 

The gradient of PQ1 , PQ2 , … will give better approximations for the gradient of the tangent line at P.

f(x) Qx  δx, y  δy 

Thus, the gradient of the curve y = f(x) at P ( x , y ) is given by

 Q1  Q2  Q3

P (x,y) δx

x

N x  δx

x

δx  f(x  δx)  f(x)  lim  lim   δx 0 δy δx 0  δx 

Tangent line : The slope of the tangent line at P

y f ( x  x )  f ( x ) lim  lim x 0 x x 0 x dy  y   lim   x 0 x dx  

dy dx

Gradient function Derived function ‘ derivative of y with respect to x’

The process of finding the derivative of a function is called DIFFERENTIATION.

DERIVATIVE OF A FUNCTION Derivative of a function f(x) at a point is given by

dy  y  f ( x   x )  f ( x )    lim    lim  dx x 0  x  x 0   x   If the limit exists. If f’(x) exists, then we say that f is differentiable at x

THE FIRST PRINCIPLE OF DERIVATIVE 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 derivatives have various applications and interpretations, including the following: 1.

Slope of tangent line For each x in the domain of f’, f’(x) is the slope of line tangent 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 t, then v =f’(t) is the velocity of the object at that time.

The steps used to find the derivative from the first principles Step 1

Find f(x) and f(x+h)

Step 2

f ( x  h)  f ( x ) Find f x   lim h0 h

Using the first principle, find the derivatives with respect to x: a)

y = 5x

b) f ( x )  x  1

Solution

(a) y = 5x;

f(x) = 5x;

f(x + h) = 5(x + h)

dy f( x  h)  f( x )  lim dx h0 h 5( x  h )  (5x )  lim h0 h 5 x  5h  5 x  lim h0 h 5h  lim h0 h  lim 5 h 0

5

Solution

b) f ( x )  x  1 , f ( x  h)  x  h  1 f( x  h)  f( x ) f ( x )  lim h0 h x  h 1  x 1  lim h 0 h  x  h 1  x 1   x  h 1  x 1   lim     h 0 h    x  h 1  x 1  ( x  h  1)  ( x  1)  lim h 0 h( x  h  1  x  1 ) 1  lim h 0 x  h  1  x  1 1  2 x 1

NOTE 1. The process of finding a derivative is called differentiation.

2. f' (x)  lim f(x  δx)  f(x) , if the limit exists. If f’(x) δx0

δx

exists for each x in the open interval (a,b), then f is said to be differentiable (or f has a derivative) over (a, b). To differentiate a function of f at x means to find its derivative at the point (x, f(x)).

3. If x0 is not in the domain of f or the limit does not exist then f is not differentiable at x0.

4. The most commonly encountered circumstances of nondifferentiability occur where the graph of f has

5.

(a)

a corner - slope of the secant lines has different limits from left and right

(b)

a vertical tangent line - slope of the secant lines has different limits from +∞ or −∞

(c)

a discontinuity

f is differentiable at a, then f is continuous at a.

d a is 1) (a )  0 constant dx

d n n1 2) ( x )  nx dx

d d 3) k f (x)  k f (x) dx dx

Note : The differentiation of a function should be with respect to the independent variable . For example , if

dy i) y  f ( x ) , then  f ' (x) dx dy ii) y  h( t ) , then  h' ( t ) dt

Differentiate the following functions with respect to x: a)

y 3 dy d  3 0 dx dx

b)

f(x) = -5x6

f’(x) = 6( 5x6-1) =  30x5

Differentiate the following functions with respect to x: c)

y2 x dy d 2 dx dx 1  2  2 1  x

d) y  1 ( x2

2x 4

dy  d  ( x 4 ) dx 2 dx

)

1 1 x2



x



1 2

 



2



2

( 4x 4 1 ) ( 4x 5 )

 2 x 5 2  5 x

Sum of function

d d d  f ( x )  g( x )   f ( x )  g( x ) dx dx dx

Difference of function

d d d  f ( x )  g( x )   f ( x )  g( x ) dx dx dx

Differentiate f  x   x3  3x 2  5x  7 with respect to x: d d d d 3 2 f ( x )  (x ) ( 3x )  (5 x )  (7 ) dx dx dx dx

Solution

 3x 2  6x  5

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) and  g( x ) du dx Thus

dy dy du  . dx du dx

Power rule – another version of chain rule

d  f(x ) dx



n

 n f ( x )

n 1

Find the differentiation of y   3x  5 

12

Solution

with respect to x:

OR

If u  3x  5 then y  u du dy 3  12u11 dx du 11

Thus,

d f ( x ) dx

dy dy du   dx du dx dy  12u11  3 dx

By using power rule,

y   3x  5 

12

dy d  (3x  5 )12 dx dx d (3x  5 ) dx  12(3x  5 )11 (3 )  12(3x  5 )12 1

 36  3x  5 

11

 36(3x  5 )11

Differentiate y 

2 3x  1

with respect to x:

Solution 1 2  y  2  3x  1 2 3x  1 1  1  1) 2 ( 3 )

dy  1  2    ( 3x dx  2 

3 2

 3(3x  1) 3  3  3x  1

Change form

Use power rule

If y = uv, where u and v are functions in terms of x, then

dy dv du u  v dx dx dx If

u y v

where u and v are functions in terms of x, then

du dv v u dy dx dx  2 dx v





Differentiate y  7x 2  3 2x3  x



with respect to x:

Solution







u  7 x2  3 du  14x dx

and

a) y  7 x 2  3 2x 3  x

dy dv du u v dx dx dx





v  2x 3  x dv  6x 2  1 dx

 



 7 x 2  3 6x 2  1  2x 3  x 14x 

2x 2

Differentiate y 

4x  3

Solution

with respect to x:

dy  dx

v

du dv u dx dx 2 v



4x  3  4x   2x 2

u  2 x2

v  4x  3   4x  3 

and

 



and

v  4x  3   4x  3 

1 2



1 dv 1 1   4x  3  2  4  dx 2

du  4x dx

 2  4x  3  2  4x  3

 

2



1  2





4x  3 4x  3



u  2x

4x  3



1 1 dv 1 2 1    4x  3  2 2 4   2  4x 2 3  2  4x  3  4x   4x dx 2 4x  3

du  4x dx

2

1 2

2 4x  3

 4x  3   4x    4x 2  4x  3

16x 2  12x  4x 2

 4x  3 

12x 2  12x

 4x  3 3

3 2



1 4x  3

Differentiate the following functions with respect to x:

d (sin x)  cos x dx d (cos x)   sin x dx d (tan x)  sec 2 x dx

a) f(x)  5 sin x  6 cos x f’ (x) = 5 cos x – 6(- sin x) = 5 cos x + 6 sin x

b) f(x)  2 tan x f’(x) = -2 sec2 x

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 2 2 du (tan (ax  b))  a sec (ax  b)  sec u dx dx

Differentiate the following with respect to x :

2 a) y  cos(5  x) 3 dy d  2  d 2  cos(5  x)   (5  x)  dx dx  3  dx 3 2  2   sin(5  x)     3  3 2 2  sin(5  x) 3 3



b) y  3 tan x   2









dy d d 2 2  3 tan x    x  dx dx dx

 x     2x   6x sec  x     3 sec

2

2

2

2



1.

2.

3.

4.

d x x (a )  a ln a dx d f(x) f(x) (a )  a ln a.f'(x) dx d x x (e )  e dx d f(x) f(x) (e )  e .f '(x) dx

Differentiate each of the following with respect to x a)

5x

TIPS

d x 5  5 x ln5 dx

b)

6cos x

use

d x (a )  a xln a dx

TIPS

d cos x d cos x 6 6 ln 6 (cos x) dx dx  6cos x ln 6  (  sin x)  6cos x ln 6 sin x

use

d f(x) (a )  af(x)ln a.f   x  dx

Differentiate each of the following with respect to x c)

e

8x

TIPS

d 8x d e  e8x  8x  dx dx e

8x

d f(x) (e )  ef(x).f '(x) use dx

  8 

 8e8x

d)

e4 sin x



d 4 sin x 4 sin x d e e e4 sin x dx dx  e4 sin x   4 cos x   4  cos x  e4 sin x



TIPS d f(x) (e )  ef(x).f '(x) use dx

dy x Find for y  2xe dx

Solution u  2x du 2 dx

ve dv  ex dx x

dy x x   2x  e  e  2  dx = 2e x  x  1

d 1 1. (ln x )  dx x d 1 f ' (x) f x   2. (ln f ( x ))  dx f x  f (x) The following laws of logarithm are used to simplify any logarithmic functions before it is differentiated.

1. lnmn   ln m  ln n m 2. ln   ln m - ln n n 3. lnm  p ln m p

Differentiate each of the following with respect to x :

a) f(x) = ln (sin x) 1 d f x   (sin x ) sin x dx cos x  sin x '

b) f(x) = ln (2x+1)3

dTIPS 1 1. (ln x )  dx x

use

d 1 f ' (x)      2 . (ln f ( x ))  f x  1. ln x   ln n f ( x ) dxmn  lnf m





m 2. ln    ln m - ln n TIPS n

Simplify use d 1 1.

(ln x ) 

3. lnm  p ln xm

use2.

p dx

d 1 f x   f ' ( x ) (ln f ( x ))  dx f x  f (x)

TIPS

c) f(x) = ln (5x-1)(3x+8) 1)

d (2x  1) dx

Simplify use d 1 (ln x ) 

= ln (5x – 1) + ln (3x + 8) 1. lnmndx  ln mx  ln n 1 1 d 1  f ( x)  (5 )  (3 ) f x   f ' ( x ) 2. (ln f ( x ))  use f x  f (x)  mdx  5x  1 3x  8 2. ln   ln m - ln n 1.

n

5 3   5x  1 3x  8

3. lnm  p ln m p

 2x  1   2  x  1

d) f(x) = ln 

TIPSd 1.

= ln (2x -1) – ln (x + 1) 2

  ln 1.e) lnmn n f(x) =m 2x3lnln 1 m x 2 . ln use    ln m - ln n3

(ln x ) 

dx Simplify

 nf ( x )

 ( 2x )

d 1 f ' (x) use      2 . (ln f ( x ))  f x  p 1 1 3   3 . ln m  p ln m  dx f x f ( x )   2 x f (x)  2 (2 )  2 ( 2x ) 3 x 1 x 1  2 2x 6x 3

TIPS

3

e) f(x) = 2x ln (3x-2)

(x2 + 1) 1 x 1 2

x

1

( 2x )

use

d ln(3x  2)  ln(3x  2) d (2x13. )dxd (ln xdy)  1x u dv  v du dx dx dx dx dx  1  d 1  2x 3  (3)  ln(3 x  2) 6 x 2 f x   f ' ( x ) 2. (ln f ( x ))  dx f x  f (x)  3x  2  6x 3   6x 2 ln( 3x  2 ) 3x  2

f ( x )  (2x 3 )

 

Let y = f(x) variable y appears alone at one side of the equations.

For example , y = 2x + 1

F(x,y) = c

y =

3x2

– 7x + 1

x 1 y x3

where c is a constant variable y

cannot be stated as a subject For example , y2 – 3yx3 = x + 1

xy + 2y = x - 1

If z =

y2,

thus

dz  2y . dy

dz But what is ? dx By using chain rule,

dy dz dz dy  2y  . dx dx dy dx

We need to use Chain Rule to find the derivative of an Implicit Function .

d [f(y)] dx

dy d = dy [ f ( y ) ] . dx

1)

d [y] = dx

2)

dy d [ y2 ] = 2y dx dx

3)

d [ y4 ] dx

dy dx

=

3y3

dy dx

dy Find for the following: dx a) x 2  y 2  25

Solution

Differentiating implicitly with respect to x

d 2 d 2 d (x )  (y )  (25 ) dx dx dx dy 2x  2y 0 dx dy 2y  2x dx dy x  dx y

x 3  y 3  3 xy  9

b)

TIPS use

Solution

dy  dy  3x  3y  3 x  y  0 dx  dx  dy 2 2 dy 3x  3y  3x  3y  0 dx dx dy 2 dy 3y  3x  3x 2  3y dx dx dy 3x  3y 2  3x 2  3y dx dy  3x 2  3y  dx 3x  3y 2 2

2





 (x 2  y)  x  y2

dy dv du u  v dx dx dx