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DERIVATIVES 
BRIDGE COURSE 
F.Y Electrical-1 
By:- 
Nisarg Amin 14BEEEG012 
Deep Pandya 14BEEEF011 
Urjit Trivedi 14BEEEG008 
Pratik Sharma 14BEEE007 
Kunj Shah 14BEEEG010
Derivatives 
The derivative, or derived function of f(x) denoted by f `(x) is defined 
as 
f x f x h f x 
= æ + - ö çè ø¸ 
`( ) lim ( ) ( ) 
0 
h 
® h 
f (x + h) - f (x) 
h 
x x + h 
P 
Q 
x 
y 
f x y dy 
= æ ö = çè ø¸ 
Leibniz Notation: `( ) lim 
h o 
d 
x dx 
® d
Differentiation from first principles 
Given that f(x) is differentiable, we can use the definition to prove 
that if f (x) = x2 then f `(5) =10 
f x = f x + h - 
f x 
`( ) lim ( ) ( ) 
0 
h 
® h 
2 2 
= + - 
lim ( ) 
h 
0 
x h x 
® h 
2 2 2 
= + + - 
lim 2 
h 
0 
x xh h x 
® h 
2 
= + 
lim 2 
h 
0 
xh h 
® h 
= + 
lim (2 ) 
h 
0 
h x h 
® h 
= lim(2 + 
) 
h 
0 
As h®0, f `(x)®2x 
 f `(5) =10 
x h 
®
NOTE Not all functions are differentiable. 
y 
y = tan(x). 
x 
Here, tan(x) has ‘breaks’ in the graph where 
the gradient is undefined 
y y = x 
Although the graph is continuous, the 
derivative at zero is undefined as the left 
derivative is negative and the right 
derivative is positive. 
x 
For a function to be differentiable, it must be continuous.
The Product Rule 
If k(x) = f (x).g(x) , then: 
k`(x) = f `(x)g(x) + g`(x) f (x) 
Using Leibniz notation, 
If y = f (x).g(x), then: 
dy df .g(x) dg . f (x) 
dx dx dx 
= + OR dy f `g g`f 
dx 
= +
1. Differentiate y = x2 sin x 
f (x) = x2 g(x) = sin x 
f `(x) = 2x g`(x) = cos x 
dy f `g g`f 
dx 
= + 
= 2xsin x + x2 cos x
The Quotient Rule 
k x f x 
If ( ) ( ) , then: 
g x 
( ) 
= 
k x = f x g x - 
g x f x 
`( ) `( ) ( ) `( ) ( ) 
2 
g x 
( ( )) 
Using Leibniz notation, 
If y f ( x 
) , then: 
g x 
( ) 
= 
df g x dg f x dy dx dx 
dx g x 
. ( ) . ( ) 
( ( )) 
2 
- 
dy = f `g - 
g`f 
dx g 
= OR 2
æ 3 
ö 
ç ¸ 
è ø 
1. Find 
d x 
dx sin 
x 
f (x) = x3 g(x) = sin x 
f `(x) = 3x2 g`(x) = cos x 
dy = f `g - 
g`f 
dx g 
2 
2 3 
x x x x 
3 sin cos 
2 
sin 
x 
= -
Sec, Cosec, cot and tan 
sec 1 the secant of 
x x 
cos 
x 
= 
cos 1 the cosecant of 
ec x x 
sin 
x 
= 
cot 1 the cotangent of 
x x 
tan 
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.
y 
x 
1 
2 
3 
4 
5 
6 
6 
5 
4 
3 
2 
1 
– 1 
– 2 
– 3 
– 4 
– 5 
– 6 
y = sec x 
123456–––––– 123456 
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 
y = cosec x 
123456–––––– 123456 
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 
y = cot x 
x x = p + np 
In general, sec is undefined for 
2 
also cosec x is undefined for x = np
x x = p + np 
In general, sec is undefined for 
2 
sec 1 
æ ö = çè ø¸ æ ö 
2 cos 
2 
p 
p 
çè ø¸ 
1 
0 
= undefined 
also cosec x is undefined for x = np
1. Find the derivative of tan x 
d ( tan x ) d sin 
x 
dx dx x 
= æ ö çè cos 
ø¸ 
(quotient rule) 
x x x x 
= - - 
cos .cos ( sin ).sin 
2 
cos 
x 
1 
cos 2 
x 
= 
= sec2 x
Exponential and Logarithmic functions 
If y = ex then dy = 
ex 
dx 
y x dy x 
If = log then = 1 ¹ 
0 e 
dx x
1. Find d e3x 
dx 
d e3x = 
3e3x 
dx 
2. Find d 
dx 
xex 
d xex ex xex 
dx 
= + = ex (1+ x) 
3. Find d ln3x 
dx 
ln3 1 .3 
d x 
dx x 
= 1 
3 
x 
=
Higher derivatives 
Function 1st derivative 2nd Derivative………..nth Derivative 
f (x) f '(x) f ''(x) f n (x) 
dy 
dx 
2 
2 
d y 
dx 
n 
n 
d y 
dx 
y 
1. y = 3x4 
dy = 
12x3 
dx 
2 
2 
d y = 
36x 
dx 
2 3 
3 d y 72x 
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. 
v dx 
= 
dt 
Acceleration is a rate of change of velocity. 
a dv 
dt 
= 
2 
2 
d x 
dt 
=
A particle travels along the x axis such that 
x(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. 
(a) When t = 0, x(0) = 5 
Hence the particle is 5m from the origin at the start of the observation. 
(b) v = dx = 12t2 - 2 
v(3) =12´9 - 2 =106ms-1 
dt 
2 
2 a d x 24t 
= = a(3) = 24´3 = 72ms-2 
dt
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
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) = 0 
2. Find critical points 
3. 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) = 0 
2. Find critical points 
3. 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 
3 2 
3 2 
of x - 6x – 36x + 2. 
f(x) = x - 6x – 36x + 2 
f’(x) = 3x - 2 
12x – 36 
f’(x) = 3(x - 2 
4x–12) 
f’(x) = 3(x-6) (x+2) 
f’(x) = 0 
3(x-6) (x+2) = 0 
x = 6 or x = -2 
Critical points = 6,-2
Now, 
f”(x) = 6x – 12 
f”(6) = 36 – 12 
= 24 > 0 
f”(-2) = -12 – 12 
= - 24 < 0 
Function has local maximum value at x = -2 
Function has local minimum value at x = 6 
Also, 
f(6) = 216 – 216 - 216 + 2 
= -214 
f(-2) = -8 – 24 + 72 + 2 
= 42 
Function has absolute maximum value at x = -2 
Function has absolute minimum value at x = 6

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Derivatives

  • 1. DERIVATIVES BRIDGE COURSE F.Y Electrical-1 By:- Nisarg Amin 14BEEEG012 Deep Pandya 14BEEEF011 Urjit Trivedi 14BEEEG008 Pratik Sharma 14BEEE007 Kunj Shah 14BEEEG010
  • 2. Derivatives The derivative, or derived function of f(x) denoted by f `(x) is defined as f x f x h f x = æ + - ö çè ø¸ `( ) lim ( ) ( ) 0 h ® h f (x + h) - f (x) h x x + h P Q x y f x y dy = æ ö = çè ø¸ Leibniz Notation: `( ) lim h o d x dx ® d
  • 3. Differentiation from first principles Given that f(x) is differentiable, we can use the definition to prove that if f (x) = x2 then f `(5) =10 f x = f x + h - f x `( ) lim ( ) ( ) 0 h ® h 2 2 = + - lim ( ) h 0 x h x ® h 2 2 2 = + + - lim 2 h 0 x xh h x ® h 2 = + lim 2 h 0 xh h ® h = + lim (2 ) h 0 h x h ® h = lim(2 + ) h 0 As h®0, f `(x)®2x f `(5) =10 x h ®
  • 4. NOTE Not all functions are differentiable. y y = tan(x). x Here, tan(x) has ‘breaks’ in the graph where the gradient is undefined y y = x Although the graph is continuous, the derivative at zero is undefined as the left derivative is negative and the right derivative is positive. x For a function to be differentiable, it must be continuous.
  • 5. The Product Rule If k(x) = f (x).g(x) , then: k`(x) = f `(x)g(x) + g`(x) f (x) Using Leibniz notation, If y = f (x).g(x), then: dy df .g(x) dg . f (x) dx dx dx = + OR dy f `g g`f dx = +
  • 6. 1. Differentiate y = x2 sin x f (x) = x2 g(x) = sin x f `(x) = 2x g`(x) = cos x dy f `g g`f dx = + = 2xsin x + x2 cos x
  • 7. The Quotient Rule k x f x If ( ) ( ) , then: g x ( ) = k x = f x g x - g x f x `( ) `( ) ( ) `( ) ( ) 2 g x ( ( )) Using Leibniz notation, If y f ( x ) , then: g x ( ) = df g x dg f x dy dx dx dx g x . ( ) . ( ) ( ( )) 2 - dy = f `g - g`f dx g = OR 2
  • 8. æ 3 ö ç ¸ è ø 1. Find d x dx sin x f (x) = x3 g(x) = sin x f `(x) = 3x2 g`(x) = cos x dy = f `g - g`f dx g 2 2 3 x x x x 3 sin cos 2 sin x = -
  • 9. Sec, Cosec, cot and tan sec 1 the secant of x x cos x = cos 1 the cosecant of ec x x sin x = cot 1 the cotangent of x x tan 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.
  • 10. y x 1 2 3 4 5 6 6 5 4 3 2 1 – 1 – 2 – 3 – 4 – 5 – 6 y = sec x 123456–––––– 123456 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 y = cosec x 123456–––––– 123456 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 y = cot x x x = p + np In general, sec is undefined for 2 also cosec x is undefined for x = np
  • 11. x x = p + np In general, sec is undefined for 2 sec 1 æ ö = çè ø¸ æ ö 2 cos 2 p p çè ø¸ 1 0 = undefined also cosec x is undefined for x = np
  • 12. 1. Find the derivative of tan x d ( tan x ) d sin x dx dx x = æ ö çè cos ø¸ (quotient rule) x x x x = - - cos .cos ( sin ).sin 2 cos x 1 cos 2 x = = sec2 x
  • 13. Exponential and Logarithmic functions If y = ex then dy = ex dx y x dy x If = log then = 1 ¹ 0 e dx x
  • 14. 1. Find d e3x dx d e3x = 3e3x dx 2. Find d dx xex d xex ex xex dx = + = ex (1+ x) 3. Find d ln3x dx ln3 1 .3 d x dx x = 1 3 x =
  • 15. Higher derivatives Function 1st derivative 2nd Derivative………..nth Derivative f (x) f '(x) f ''(x) f n (x) dy dx 2 2 d y dx n n d y dx y 1. y = 3x4 dy = 12x3 dx 2 2 d y = 36x dx 2 3 3 d y 72x dx = etc. etc. etc.
  • 16. 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.
  • 17. 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.
  • 18. Applications of Derivatives Let displacement from an origin be a function of time. x = f (t) Velocity is a rate of change of displacement. v dx = dt Acceleration is a rate of change of velocity. a dv dt = 2 2 d x dt =
  • 19. A particle travels along the x axis such that x(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. (a) When t = 0, x(0) = 5 Hence the particle is 5m from the origin at the start of the observation. (b) v = dx = 12t2 - 2 v(3) =12´9 - 2 =106ms-1 dt 2 2 a d x 24t = = a(3) = 24´3 = 72ms-2 dt
  • 20. 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.
  • 21. Local max Local max Local min Local & Absolute Max Local & Absolute
  • 22. 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) = 0 2. Find critical points 3. If f”(x) > 0 then function has local minimum value at ‘x’. If f”(x) < 0 then function has local maximum value at ‘x’.
  • 23. 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) = 0 2. Find critical points 3. 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.
  • 24. Illustration :- • Find local and absolute maximum and minimum 3 2 3 2 of x - 6x – 36x + 2. f(x) = x - 6x – 36x + 2 f’(x) = 3x - 2 12x – 36 f’(x) = 3(x - 2 4x–12) f’(x) = 3(x-6) (x+2) f’(x) = 0 3(x-6) (x+2) = 0 x = 6 or x = -2 Critical points = 6,-2
  • 25. Now, f”(x) = 6x – 12 f”(6) = 36 – 12 = 24 > 0 f”(-2) = -12 – 12 = - 24 < 0 Function has local maximum value at x = -2 Function has local minimum value at x = 6 Also, f(6) = 216 – 216 - 216 + 2 = -214 f(-2) = -8 – 24 + 72 + 2 = 42 Function has absolute maximum value at x = -2 Function has absolute minimum value at x = 6