DERIVATIVESBRIDGE COURSE

F.Y Electrical-1By:-Nisarg Amin 14BEEEG012Deep Pandya 14BEEEF011Urjit Trivedi 14BEEEG008Pratik Sharma 14BEEE007Kunj Shah 14BEEEG010

DerivativesThe derivative, or derived function of f(x) denoted by f `(x) is defined as

0

( ) ( )`( ) lim

h

f x h f xf x

h

( ) ( )f x h f x

h

x x + h

P

Q

x

y

Leibniz Notation: `( ) limh o

y dyf x

x dx

Differentiation from first principlesGiven that f(x) is differentiable, we can use the definition to prove that if 2( ) then `(5) 10f x x f

0

( ) ( )`( ) lim

h

f x h f xf x

h

2 2

0

( )limh

x h x

h

2 2 2

0

2limh

x xh h x

h

2

0

2limh

xh h

h

0

(2 )limh

h x h

h

0lim(2 )h

x h

As 0, `( ) 2h f x x

`(5) 10f

NOTE Not all functions are differentiable.

y

x

y = tan(x).

Here, tan(x) has ‘breaks’ in the graph where the gradient is undefined

Although the graph is continuous, the derivative at zero is undefined as the left derivative is negative and the right derivative is positive.

x

y y x

For a function to be differentiable, it must be continuous.

The Product RuleIf ( ) ( ). ( ) , then:k x f x g x

`( ) `( ) ( ) `( ) ( )k x f x g x g x f x

Using Leibniz notation,

If ( ). ( ) , then:y f x g x

. ( ) . ( )dy df dg

g x f xdx dx dx

OR ` `dy

f g g fdx

21. Differentiate siny x x

2( )f x x ( ) sing x x

`( ) 2f x x `( ) cosg x x

` `dy

f g g fdx

22 sin cosx x x x

The Quotient Rule( )

If ( ) , then:( )

f xk x

g x

2

`( ) ( ) `( ) ( )`( )

( ( ))

f x g x g x f xk x

g x

Using Leibniz notation,

( )If , then:

( )

f xy

g x

2

. ( ) . ( )

( ( ))

df dgg x f xdy dx dx

dx g x

OR

2

` `dy f g g f

dx g

3

1. Find sin

d x

dx x

3( )f x x ( ) sing x x

2`( ) 3f x x `( ) cosg x x

2

` `dy f g g f

dx g

2 3

2

3 sin cos

sin

x x x x

x

Sec, Cosec, cot and tan1

sec the secant of cos

x xx

1cos the cosecant of

sinec x x

x

1cot the cotangent of

tanx x

x

Unlike the sine and cosine functions, the graphs of sec and cosec functions have ‘breaks’ in them.

The functions are otherwise continuous but for certain values of x, are undefined.

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

y

x

1

1

2

2

3

3

4

4

5

5

6

6

1

1

2

2

3

3

4

4

5

5

6

6

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

– 6

– 6

secy x

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

y

x

1

1

2

2

3

3

4

4

5

5

6

6

1

1

2

2

3

3

4

4

5

5

6

6

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

– 6

– 6

cosec y x

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

y

x

1

1

2

2

3

3

4

4

5

5

6

6

1

1

2

2

3

3

4

4

5

5

6

6

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

– 6

– 6

coty x

In general, sec is undefined for 2

x x n

also cosec is undefined for x x n

In general, sec is undefined for 2

x x n

1sec

2 cos2

1

0 undefined

also cosec is undefined for x x n

1. Find the derivative of tan x

sintan

cos

d d xx

dx dx x

(quotient rule)

2

cos .cos ( sin ).sin

cos

x x x x

x

2

1

cos x

2sec x

Exponential and Logarithmic functions

If then x xdyy e e

dx

1If log then 0e

dyy x x

dx x

31. Find xde

dx

3 33x xde e

dx

d2. Find

dxxxe

x x xdxe e xe

dx (1 )xe x

3. Find ln3d

xdx

1ln3 .3

3

dx

dx x

1

x

Higher derivativesFunction 1st derivative 2nd Derivative………..nth Derivative

( )f x '( )f x ''( )f x ( )nf x

dy

dx

2

2

d y

dx

n

n

d y

dxy

41. 3y x312

dyx

dx

22

236

d yx

dx

3

372

d yx

dx etc. etc. etc.

The second derivative, or second order derivative, is the derivative of the derivative of a function. The derivative of the function f(x) may be denoted by f’(x) , and its double (or "second") derivative is denoted by f’’(x) . This is read as "f double prime of x," or "The second derivative of f(x)." Because the derivative of function  is defined as a function representing the slope of function , the double derivative is the function representing the slope of the first derivative function.

Furthermore, the third derivative is the derivative of the derivative of the derivative of a function, which can be represented by f’’’(x) . This is read as "f triple prime of x", or "The third derivative of f(x)". This can continue as long as the resulting derivative is itself differentiable, with the fourth derivative, the fifth derivative, and so on. Any derivative beyond the first derivative can be referred to as a higher order derivative.

Applications of Derivatives

Let displacement from an origin be a function of time.

( )x f t

Velocity is a rate of change of displacement.

dxv

dt

Acceleration is a rate of change of velocity.

dva

dt

2

2

d x

dt

A particle travels along the x axis such thatx(t) = 4t3 – 2t + 5,where x represents its displacement in metres from the origin ‘t’

seconds after observation began.

(a) How far from the origin is the particle at the start of observation?

(b) Calculate the velocity and acceleration of the particle after 3 seconds.

( ) When 0, (0) 5a t x

Hence the particle is 5m from the origin at the start of the observation.

2( ) 12 2dx

b v tdt

1(3) 12 9 2 106v ms

2

224

d xa t

dt 2(3) 24 3 72a ms

Maximum And Minimum

Two types of maxima and minima:

1) Local maximum and minimum

2) Absolute maximum and minimum

Maximum and Minimum are collectively called EXTREMA.

Local max

Local max

Local min

Local & Absolute Max

Local & Absolute Min

Local Maximum and Minimum

Local extrema are the extrema which occur in the neighborhood of the function.

• How to find local maximum and minimum?Let f(x) be any function.1. Assume f(x) = 02. Find critical points3. If f”(x) > 0 then function has local minimum

value at ‘x’. If f”(x) < 0 then function has local maximum

value at ‘x’.

Absolute Maximum and Minimum

Absolute extrema are the largest and smallest values that a function takes on over its entire domain.

• How to find absolute maximum and minimum? Let f(x) be any function.1. Assume f(x) = 02. Find critical points3. Put critical points in f(x). The value of ‘x’ for which maximum value of f(x) is

obtained is Absolute Maximum. The value of ‘x’ for which minimum value of f(x) is

obtained is Absolute Minimum.

Illustration :-

• Find local and absolute maximum and minimum of x - 6x – 36x + 2.

f(x) = x - 6x – 36x + 2f’(x) = 3x - 12x – 36f’(x) = 3(x - 4x–12)f’(x) = 3(x-6) (x+2)f’(x) = 03(x-6) (x+2) = 0 x = 6 or x = -2Critical points = 6,-2

3 2

3 2

2

2

Now, f”(x) = 6x – 12 f”(6) = 36 – 12 = 24 > 0 f”(-2) = -12 – 12 = - 24 < 0Function has local maximum value at x = -2Function has local minimum value at x = 6Also, f(6) = 216 – 216 - 216 + 2 = -214 f(-2) = -8 – 24 + 72 + 2 = 42Function has absolute maximum value at x = -2Function has absolute minimum value at x = 6